Testing hypothesis (methods of testing the statement of organizations)

My ppt is about the testing hypothesis which is used in statistics to check whether the statement of company, organization, or institution is true or false

1. ( Directional Hypothesis and Non- Directional Hypothesis )

2. Example: Let’s say we need to determine if girls on average score higher than 600 in the exam. We have the information that the standard deviation for girls’ scores is 100. So, we collect the data of 20 girls by using random samples and record their marks. we also set our ⍺ value (significance level) to be 0.05. Since the P-value is less than 0.05, we can reject the null hypothesis  and conclude based on our result that Girls on average scored higher than 600.

3. let’s say we want to know if Girls on average score 10 marks more than the boys. We have the information that the standard deviation for girls’ Score is 100 and for boys’ score is 90. Then we collect the data of 20 girls and 20 boys by using random samples and record their marks. we also set our ⍺ value (significance level) to be 0.05. Thus, we can  conclude based on the P-value that we fail to reject the Null Hypothesi s. We don’t have enough evidence to conclude that girls on average score of 10 marks more than the boys. Pretty simple, right?

4. Let’s say we want to determine if on average girls score more than 600 in the exam. We do not have the information related to variance (or standard deviation) for girls’ scores. To a perform t-test, we randomly collect the data of 10 girls with their marks and choose our ⍺ value (significance level) to be 0.05 for Hypothesis Testing. Our  P-value is greater than 0.05 thus we fail to reject the null hypothesis  and don’t have enough evidence to support the hypothesis that on average, girls score more than 600 in the exam.

5. let’s say we want to determine if on average, boys score 15 marks more than girls in the exam. We do not have the information related to variance (or standard deviation) for girls’ scores or boys’ scores. To perform a t-test. we randomly collect the data of 10 girls and boys with their marks. We choose our ⍺ value (significance level) to be 0.05 as the criteria for Hypothesis Testing. Thus,  P-value is less than 0.05 so we can reject the null hypothesis  and conclude that on average boys score 15 marks more than girls in the exam.

6. The Student’s t-Distribution and Statistical Inference: The test is used to compare samples from two different batches. t test is a useful technique for comparing mean values of two sets of numbers. • The comparison will provide you with a statistic for evaluating whether the difference between two means is statistically significant. Types: 1. The one-sample t test is used compare a single sample with a population value. For example, a test could be conducted to compare the average salary of nurses within a company with a value that was known to represent the national average for nurses. 2. The independent-sample t test is used to compare two groups ' scores on the same variable. For example, it could be used to compare the salaries of nurses and physicians to evaluate whether there is a difference in their salaries. 3. The paired-sample t test is used to compare the means of two variables within a single group. For example, it could be used to see if there is a statistically significant difference between starting salaries and current salaries among the general nurses in an organization. Conditions/Assumptions Case I N<30 Population is approximately normal Case II N ≥ 30

7. Assumptions of t-Test • Dependent variables are interval or ratio. • The population from which samples are drawn is normally distributed. • Samples are randomly selected. • The groups have equal variance (Homogeneity of variance). • The t-statistic is robust (it is reasonably reliable even if assumptions are not fully met . The one-sample t test With d.f ʋ = n-1 The independent-sample t test: , when but unknown. d.f = n1 + n2 - 2 with d.f. when and unknown . The paired-sample t test: Where, ,   ***95% Confidence Interval Estimate of Mean  

8. Q: A psychologist claims that the mean age at which children start walking is 12.5 months. Carol wanted to check if this claim is true. She took a random sample of 18 children and found that the mean age at which these children started walking was 12.9 months with a standard deviation of .80 month. It is known that the ages at which all children start walking are approximately normally distributed. Find the p -value for the test that the mean age at which all children start walking is different from 12.5 months. What will your conclusion be if the significance level is 1 %? Q: A random sample of 14 observations taken from a population that is normally distributed produced a sample mean of 212.37 and a standard deviation of 16.35. Find the critical and observed values of t and the range for the p -value for each of the following tests of hypotheses, using α = .10. a. H 0: μ = 205 versus H 1: μ ≠ 205 b. H 0: μ = 205 versus H 1: μ > 205 Q: A random sample of 8 observations taken from a population that is normally distributed produced a sample mean of 44.98 and a standard deviation of 6.77. Find the critical and observed values of t and the range for the p -value for each of the following tests of hypotheses, using α = .05. a. H 0: μ = 50 versus H 1: μ ≠ 50 b. H 0: μ = 50 versus H 1: μ < 50 Q: Consider the null hypothesis H 0: μ = 100. Suppose that a random sample of 35 observations is taken from this population to perform this test. Using a significance level of .01, show the rejection and non-rejection regions and find the critical value(s) of t when the alternative hypothesis is as follows. a. H 1: μ ≠ 100 b. H 1: μ > 100 c. H 1: μ < 100 “One sample t-test”

9. Q: A sample of 14 cans of Brand I diet soda gave the mean number of calories of 23 per can with a standard deviation of 3 calories. Another sample of 16 cans of Brand II diet soda gave the mean number of calories of 25 per can with a standard deviation of 4 calories. At a 1% significance level, can you conclude that the mean numbers of calories per can are different for these two brands of diet soda? Assume that the calories per can of diet soda are approximately normally distributed for each of the two brands and that the standard deviations for the two populations are equal. Brand I diet soda: n1 = 14, s1= 3 Brand II diet soda: n2 = 16, s2=4 H 0: μ 1 - μ 2 = 0 (The mean number of calories are not different) H 1: μ 1 - μ 2 ≠ 0 (The mean number of calories are different) Degrees of freedom = n 1 + n 2 - 2 = 14 + 16 - 2 = 28 Sp = 3.57, t- cal =-1.531 Q: A sample of 40 children from New York State showed that the mean time they spend watching television is 28.50 hours per week with a standard deviation of 4 hours. Another sample of 35 children from California showed that the mean time spent by them watching television is 23.25 hours per week with a standard deviation of 5 hours. Using a 2.5% significance level, can you conclude that the mean time spent watching television by children in New York State is greater than that for children in California? Assume that the standard deviations for the two populations are equal. “Two sample t-test when variances are equal but unknown”  

10. Q: The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. n1 = 21 x1 = 13.97 s1 = 3.78 n2 = 20 x2 = 15.55 s2 = 3.26 Test at a 5% significance level if the two population means are different. Q: The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. n1 = 55 x1 = 90.40 s1 = 11.60 n2 = 50 x2 = 86.30 s2 = 10.25 Test at a 1% significance level if the two population means are different. Q: The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. n1 = 55 x1 = 90.40 s1 = 11.60 n2 = 50 x1 = 86.30 s2 = 10.25 Test at a 5% significance level if μ1 is greater than μ2 . “Two sample t-test when variances are equal but unknown”

11. Q: Assuming that the two populations are normally distributed with unequal and unknown population standard deviations, construct a 95% confidence interval for μ1 − μ2 for the following. n1 = 14 x1 = 109.43 s1 = 2.26 n2 = 15 x2 = 113.88 s2 = 5.84 Q: Assuming that the two populations have unequal and unknown population standard deviations, construct a 99% confidence interval for μ1 − μ2 for the following. n1 = 48 x1 = .863 s1 = .176 n2 = 46 x2 = .796 s2 = .068 Q: The following information was obtained from two independent samples selected from two normally distributed populations with unequal and unknown population standard deviations. n1 = 14 x1 = 109.43 s1 = 2.26 n2 = 15 x2 = 113.88 s2 = 5.84 Test at a 5% significance level if the two population means are different. Q: The following information was obtained from two independent samples selected from two populations with unequal and unknown population standard deviations. n1 = 48 x1 = .863 s1 = .176 n2 = 46 x2 = .796 s2 = .068 Test at a 1% significance level if the two population means are different. “Two sample t-test when variances are unequal and unknown”

12. Paired or Matched Samples Two samples are said to be paired or matched samples when for each data value collected from one sample there is a corresponding data value collected from the second sample, and both these data values are collected from the same source. A researcher wanted to find the effect of a special diet on systolic blood pressure. She selected a sample of seven adults and put them on this dietary plan for 3 months. The following table gives the systolic blood pressures (in mm Hg) of these seven adults before and after the completion of this plan. Subject: 1 2 3 4 5 6 7 Before: 210 180 195 220 231 199 224 After : 193 186 186 223 220 183 233 Let μd be the mean reduction in the systolic blood pressures due to this special dietary plan for the population of all adults. Construct a 95% confidence interval for μd . Assume that the population of paired differences is approximately normally distributed . Q: Find the following confidence intervals for μd , assuming that the populations of paired differences are normally distributed. a. n = 11 d = 25.4, sd = 13.5, confidence level = 99% b. n = 23, d = 13.2, sd = 4.8, confidence level = 95% c. n = 18, d = 34.6, sd = 11.7, confidence level = 90% Q: Perform the following tests of hypotheses, assuming that the populations of paired differences are normally distributed. a. H 0: μd = 0, H 1: μd ≠ 0, n = 9, d = 6.7, sd = 2.5, α = .10 b. H 0: μd = 0, H 1: μd > 0, n = 22, d = 14.8, sd = 6.4, α = .05 c. H 0: μd = 0, H 1: μd < 0, n = 17, d = -9.3, sd = 4.8, α = .01

13. Type I and Type II Errors 1.Type I error refers to the situation when we reject the null hypothesis when it is true (H0 is wrongly rejected). H0 : there is no difference between the two drugs on average. Type I error will occur if we conclude that the two drugs produce different effects when actually there isn’t a difference Prob (Type I error) = significance level = α A Type I error occurs when a true null hypothesis is rejected. The value of α represents the probability of committing this type of error; that is, α = P ( H 0 is rejected ⃒ H 0 is true) The value of α represents the significance level of the test. 2 . Type II error refers to the situation when we accept the null hypothesis when it is false. H0: there is no difference between the two drugs on average. Type II error will occur if we conclude that the two drugs produce the same effect when actually there is a difference. Prob (Type II error) = ß A Type II error occurs when a false null hypothesis is not rejected. The value of β represents the probability of committing a Type II error; that is, β = P ( H 0 is not rejected ∣ H 0 is false) The value of 1 - β is called the power of the test . It represents the probability of rejecting H 0 when it is false.

14. One Sample t-Tests One sample t-tests can be used to determine if the mean of a sample is different from a particular value. In this example, we will determine if the mean number of older siblings that the PSY 216 students have is greater than 1. We will follow our customary steps: Write the null and alternative hypotheses first: H : µ 216 Students  ≤ 1 H 1 : µ 216 Students  > 1 Where µ is the mean number of older siblings that the PSY 216 students have. Determine if this is a one-tailed or a two-tailed test. Because the hypothesis involves the phrase "greater than", this must be a one tailed test. Specify the α level: α = .05 Determine the appropriate statistical test. The variable of interest, older, is on a ratio scale, so a z-score test or a t-test might be appropriate. We will use the t-test instead. Calculate the t value .

15. Determine if we can reject the null hypothesis or not. The decision rule is: if the one-tailed critical t value is less than the observed t AND the means are in the right order, then we can reject H . In this example, the critical t is 1.679 (from the table of critical t values) and the observed t is 1.614, so we fail to reject H . That is, there is insufficient evidence to conclude that the mean number of older siblings for the PSY 216 classes is larger than 1 . As P-value is 0.115 is greater than 0.05. we give the same decision.

16. The Independent Samples t-test: The Independent Samples t-test can be used to see if two means are different from each other when the two samples that the means are based on were taken from different individuals who have not been matched. In this example, we will determine if the students who intend to earn a PhD and who don’t intend to earn PhD of PSY 216 have a different number of older siblings. We will follow our customary steps: Write the null and alternative hypotheses first: H : µ PhD (Yes)  = µ PhD ( No) H 1 : µ PhD (Yes )  ≠ µ PhD (No) Where µ is the mean number of older siblings that the PSY 216 students have. Determine if this is a one-tailed or a two-tailed test. Because the hypothesis involves the phrase "different" and no ordering of the means is specified, this must be a two tailed test. Specify the α level: α = .05 Determine the appropriate statistical test. The variable of interest, older, is on a ratio scale, so a z-score test or a t-test might be appropriate. Because the population standard deviation is not known, the z-test would be inappropriate. Furthermore, there are different students in sections 1 and 2 of PSY 216, and they have not been matched. Because of these factors, we will use the independent samples t-test.

17. Independent Samples t-Tests Cut Point Groups: Sometimes you want to perform a t-test but the groups are defined by a variable that is not dichotomous (i.e., it has more than two values.) We may want to see if the number of older siblings is different for students who have higher GPAs than for students who have lower GPAs. Since there is no single value of GPA that specifies "higher" or "lower", we cannot proceed exactly as we did before. Before proceeding, decide which value you will use to divide the GPAs into the higher and lower groups . H : µ lower GPA  = µ higher GPA H 1 : µ lower GPA  ≠ µ Higher GPA Where µ is the mean number of older siblings that the PSY 216 students have. Determine if this is a one-tailed or a two-tailed test. Because the hypothesis involves the phrase "different" and no ordering of the means is specified, this must be a two tailed test. Specify the α level: α = .05 Determine the appropriate statistical test. The variable of interest, older, is on a ratio scale, so a z-score test or a t-test might be appropriate. Different students have higher and lower GPAs, so we have a between-subjects design. Because of these factors, we will use the independent samples t-test .

18. " Levene's Test for Equality of Variances" tell us whether an assumption of the t-test has been met. The significance (p value) of Levene's test is .0.149. If this value is less than or equal to α level for this test, we can reject the null hypothesis that the variabilities of the two groups are equal, implying that the variances are unequal. In this example, 0.149 is larger than our α level of .05, so we will assume that the variances are equal and we will use the middle row of the output. The column labeled "t" gives the observed or calculated t value. In this example, assuming equal variances, the t value is 1.795. (We can ignore the sign of t when using a two-tailed t-test.) The column labeled "df" gives the degrees of freedom associated with the t test. In this example, there are 36 degrees of freedom. The column labeled "Sig. (2-tailed)" gives the two-tailed p value associated with the test. In this example, the p value is 0.056. If this had been a one-tailed test, we would need to look up the critical t in a table. Decide if we can reject H : As before, the decision rule is given by: If p ≤ α , then reject H . In this example, 0.056 is greater than .05, so we fail to reject H . That implies that there is not sufficient evidence to conclude that people with higher or lower GPAs have different number of older siblings .

19. Paired Samples t-Tests: When two samples are involved and the values for each sample are collected from the same individuals (that is, each individual gives us two values, one for each of the two groups), or the samples come from matched pairs of individuals then a paired-samples t-test may be an appropriate statistic to use. The paired samples t-test can be used to determine if two means are different from each other when the two samples that the means are based on were taken from the matched individuals or the same individuals. We will determine if the students have different numbers of younger and older siblings. H : µ older  = µ younger H 1 : µ older  ≠ µ younger Where µ is the mean number of siblings that the PSY 216 students have. Determine if this is a one-tailed or a two-tailed test. Because the hypothesis involves the phrase "different" and no ordering of the means is specified, this must be a two tailed test. Specify the α level: α = .05 Determine the appropriate statistical test. The variables of interest, older and younger, are on a ratio scale, so a z-score test or a t-test might be appropriate. Because the population standard deviation is not known, the z-test would be inappropriate. Furthermore, the same students are reporting the number of older and younger siblings, we have a within-subjects design. Because of these factors, we will use the paired samples t-test.

20. The column labeled "t" gives the observed or calculated t value. In this example, the t value is 1.171 (you can ignore the sign.) The column labeled "df" gives the degrees of freedom associated with the t test. In this example, there are 37 degrees of freedom. The column labeled "Sig. (2-tailed)" gives the two-tailed p value associated with the test. In this example, the p value is .249. If this had been a one-tailed test, we would need to look up the critical value of t in a table. Decide if we can reject H : As before, the decision rule is given by : If p ≤ α, then reject H . In this example, .249 is not less than or equal to .05, so we fail to reject H . That implies that there is insufficient evidence to conclude that the number of older and younger siblings is different.