Question

In today’s “wired” society, students believe they can multitask. Research suggests that 5% of individuals truly have the ability to multitask. Researchers are interested in finding out if students can multitask or if they perform worse while multitasking. In an introductory statistics class, students were randomly divided into two groups: texting group and cell phone off group. 92 students were required to send three text messages during the lecture and 92 students were required to turn off their cell phones. At the end of class, a quiz was administered based on information shared during class. The mean mark (in %) for the texting group was 52.18 with a standard deviation of 9.91. The mean mark (in %) for the cell phone off group was 68.76 with a standard deviation of 10.42.

a) Identify whether the sampling method used in this study is independent or dependent. Explain.

b) Estimate the mean difference in scores between the texting group and cell phone off group with 90% confidence. Does the sample evidence suggest there is a difference in mean marks between the groups? If so, interpret the difference by identifying which group performs worse.

c) Conduct a hypothesis test to determine whether the data suggests students score worse in the texting group at the ? = 0.05 level of significance. Use the critical value method. (Show all six steps for hypothesis testing.)

Please answer step-by-step showing the formulas for the calculations in part B and part C, I need clarity on how to do each step. The charts last time were not helpful.

Answer 1

A) THE SAMPLING METHOD USED IN THIS STUDY IS INDEPENDENT. BECAUSE BOTH GROUPS ARE ASSIGNED TO DIFFERENT TASK..

To proceed further let us check the equality of variances assumption.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

9.912 10.499 = 0.905

The critical values are FL​=0.661 and FU​=1.512, and since F=0.905, then the null hypothesis of equal variances is not rejected.

b) Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 18817.39 / 182 = 103.39

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((103.39/92) + (103.39/92)) = 1.5

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = -16.58 ± (1.65 * 1.5) = -16.58 ± 2.475

-16.58, 90% CI [-19.055,-14.105].

You can be 90% confident that the difference between your two population means (μ₁ - μ₂) lies between -19.055 and -14.105

c) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=182. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=1.973, for α=0.05 and df=182.

The rejection region for this two-tailed test is R={t:∣t∣>1.973}.

(3) Test Statistics

X1 - X2 (n-1) s +(12-1) 11-12- 2 (1 + 1) V 52.18 - 68.76 = -1.088 (92-19.91° +(92-1)10.42-1 92-92-2 ( + 1 ) )

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=1.088≤tc​=1.973, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.2782 , and since p=0.2782≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.