Question

(1 point) Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to their students. A summary of the class sizes, class means, and standard deviations is given below: ni 57, T = 86.4, S1 17.5 n2 = 60, T2 = 81.5, S2 = 18.1 Is there evidence, at an a = 0.05 level of significance, to conclude that there is a difference in the two classes? Carry out an appropriate hypothesis test, filling in the information requested. A. The value of the standardized test statistic: 1.487 B. The p-value is 1396 C. Your decision for the hypothesis test: A. Reject H. B. Reject H C. Do Not Reject Ho. D. Do Not Reject H.

Answer 1

Solution:

The provided sample means are shown below:

Xˉ1​=86.4 Xˉ2​=81.5

Also, the provided population standard deviations are:

σ1​=17.5 σ2​=18.1

and the sample sizes are n1​=57 and n2​=60.

(1) Null and Alternative Hypotheses

Claim: Their is difference between this two classes of statistics i.e Ha:μ1​̸​=μ2​

The following null and alternative hypotheses need to be tested:

Ho:μ1​=μ2​

Ha:μ1​̸​=μ2​ ...(claim)

This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is

zc​=1.96.

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

I-X Voni + on 86.4 - 81.5 17.52/57+18.12/60 = 1.489

(4) Decision about the null hypothesis

Since it is observed that ∣z∣=1.489≤zc​=1.96, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1366, and since p=0.1366≥0.05, it is concluded that the null hypothesis is not rejected.

Confidence Interval

The 95% confidence interval for μ1​−μ2​ is −1.551<μ1​−μ2​<11.351.

Graphically

p-value = Test 0.1366 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Hypothesis Test :

Ho:μ1​=μ2​

Ha:μ1​̸​=μ2​

Test Statistics:

standardised test statistic z=1.489

P value:

The p-value is p=0.1366

Conclusion:

Do not reject H0

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