Question

A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. n Proctored Nonproctorec 12 31 30 75.98 86.26 10.57 19.42 X in a. Test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests What are the null and alternative hypotheses? O A. Ho H 742 H₂ H <H2 O C. Ho H = 12 H, H1 H2 OB. Ho H1 - H2 H₂ H > H₂ OD. H H = H2 H, H. <H2

Answer 1

Sol:

Two tailed test

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Proctered = u_{Non Proctered}
Alternative hypothesis: u_Proctered$\neq$ u_{Non Proctered}

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = 3.76448
DF = 67
t = [ (x₁ - x₂) - d ] / SE

NN S.e = + 1 2

t = - 2.13

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 67 degrees of freedom is more extreme than -2.13; that is, less than -2.13 or greater than 2.13.

P-value = P(t < - 2.13) + P(t > 2.13)

Use the t-value calculator for finding p-values.

P-value = 0.018 + 0.018

P-value = 0.036

Interpret results. Since the P-value (0.036) is greater than the significance level (0.01), we failed to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that mean of proctered tests and non-proctered tests is same.

One tailed test

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Proctered = u_{Non Proctered}
Alternative hypothesis: u_Proctered < u_{Non Proctered}

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = 3.76448
DF = 67
t = [ (x₁ - x₂) - d ] / SE

NN S.e = + 1 2

t = - 2.13

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesised difference between the population means, and SE is the standard error.

Since we have a one-tailed test, the P-value is the probability that a t statistic having 67 degrees of freedom is less than -2.13.

P-value = P(t < - 2.13)

Use the t-value calculator for finding p-values.

P-value = 0.018

Interpret results. Since the P-value (0.018) is greater than the significance level (0.01), we failed to reject the null hypothesis.