We are given three different hypothesis testing scenarios and we need to determine whether to reject or fail to reject the null hypothesis in each case. a. We have a t-statistic $t(50) = 2.35$, a significance level $\alpha = 0.01$, and it's a one-tailed test to the right. b. We have sample means $\bar{X_1} = 54$ and $\bar{X_2} = 44$, sample sizes $n_1 = 14$ and $n_2 = 10$, the standard error of the difference in means $s_{\bar{X_1} - \bar{X_2}} = 9.85$, and a significance level $\alpha = 0.05$, for a two-tailed test. c. We have a 95% confidence interval of $(-0.50, 2.10)$.

1. Problem Description

We are given three different hypothesis testing scenarios and we need to determine whether to reject or fail to reject the null hypothesis in each case.

a. We have a t-statistic $t(50) = 2.35$, a significance level $\alpha = 0.01$, and it's a one-tailed test to the right.

b. We have sample means $\bar{X_1} = 54$ and $\bar{X_2} = 44$, sample sizes $n_1 = 14$ and $n_2 = 10$, the standard error of the difference in means $s_{\bar{X_1} - \bar{X_2}} = 9.85$, and a significance level $\alpha = 0.05$, for a two-tailed test.

c. We have a 95% confidence interval of $(-0.50, 2.10)$.

2. Solution Steps

a. One-tailed t-test to the right:

We have $t(50) = 2.35$ and $\alpha = 0.01$. We need to find the critical t-value for a one-tailed test with 50 degrees of freedom at $\alpha = 0.01$. Using a t-table or calculator, we find that the critical t-value, $t_{0.01, 50}$, is approximately $2.403$. Since our test statistic ($2.35$) is less than the critical value ($2.403$), we fail to reject the null hypothesis.

b. Two-tailed test for the difference of means:

The null hypothesis is that the difference in means is zero, i.e., $H_0: \mu_1 = \mu_2$.

The alternative hypothesis is that the difference in means is not zero, i.e., $H_1: \mu_1 \ne \mu_2$.

We are given $\bar{X_1} = 54$, $\bar{X_2} = 44$, $n_1 = 14$, $n_2 = 10$, $s_{\bar{X_1} - \bar{X_2}} = 9.85$, and $\alpha = 0.05$.

The test statistic is:

$t = \frac{\bar{X_1} - \bar{X_2}}{s_{\bar{X_1} - \bar{X_2}}} = \frac{54 - 44}{9.85} = \frac{10}{9.85} \approx 1.015$

The degrees of freedom for this test can be approximated using Welch's t-test formula, but for simplicity and considering that we are given $s_{\bar{X_1} - \bar{X_2}}$, we can assume that the degrees of freedom have already been accounted for. The total number of observations is $n_1 + n_2 = 14+10 = 24$. If we assume that the degrees of freedom is close to $24-2=22$, or even closer to $min(n_1-1,n_2-1) = 9$, but without more information, it would not change drastically. Looking up the critical t-value for a two-tailed test with $\alpha = 0.05$ and assuming the degrees of freedom is 9 (the minimum of the sample sizes minus one) in a t-table we get approximately $t_{0.025, 9} = 2.262$. Since the absolute value of our test statistic (1.015) is less than the critical value (2.262), we fail to reject the null hypothesis.

c. Confidence Interval:

We have a 95% confidence interval of $(-0.50, 2.10)$. Since the null hypothesis for a test of the difference between two means is that the difference is zero ($\mu_1 - \mu_2 = 0$), we check if zero is within the confidence interval. Because $0$ lies within the interval $(-0.50, 2.10)$, we fail to reject the null hypothesis at $\alpha = 0.05$.

3. Final Answer

a. Fail to reject the null hypothesis.

b. Fail to reject the null hypothesis.

c. Fail to reject the null hypothesis.