(a) A statistician randomly sampled 100 observations and found X̄ = 106 and s = 35. Calculate the t statistic and p value for testing.     H0 : μ = 100 against HA : μ > 100.Carry out the test at the 1% level of significance.(b) Repeat part (a), with s = 25.(c) Repeat part (a), with s = 15.

In carrying out a one-sample t-test for a mean, the hypotheses H0:mu=10 against HA:mu>10 were tested.A sample of size 21 was used, giving a t-statistic of 2.00.Determine the P-value associated with this test.

<p>a) To infer if the population mean is less than 100 at the 10% significance level, we can perform a one-sample t-test.</p> <p>Step 1: State the null hypothesis (H0) and the alternative hypothesis (H1). H0: μ = 100 (The population mean is equal to 100) H1: μ &lt; 100 (The population mean is less than 100)</p> <p>Step 2: Calculate the test statistic. The formula for the t statistic is:</p> <p>t = (x̄ - μ) / (s/√n)</p> <p>where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.</p> <p>t = (75 - 100) / (50/√8) = -1.414</p> <p>Step 3: Determine the critical value from the t-distribution table. For a one-tailed test at the 10% significance level with 7 degrees of freedom (n-1), the critical value is -1.415.</p> <p>Step 4: Compare the test statistic with the critical value. If the test statistic is less than the critical value, we reject the null hypothesis. In this case, -1.414 is greater than -1.415, so we do not reject the null hypothesis. Therefore, we cannot infer at the 10% significance level that the population mean is less than 100.</p> <p>b) If we know that the population standard deviation is 50, we can perform a z-test instead of a t-test.</p> <p>Step 1: The null and alternative hypotheses are the same as in part (a).</p> <p>Step 2: Calculate the z statistic using the formula:</p> <p>z = (x̄ - μ) / (σ/√n)</p> <p>where σ is the population standard deviation.</p> <p>z = (75 - 100) / (50/√8) = -1.414</p> <p>Step 3: Determine the critical value from the z-distribution table. For a one-tailed test at the 10% significance level, the critical value is -1.28.</p> <p>Step 4: Compare the z statistic with the critical value. In this case, -1.414 is less than -1.28, so we reject the null hypothesis. Therefore, we can infer at the 10% significance level that the population mean is less than 100.</p> <p>c) The test statistics differ because a t-test is used when the population standard deviation is unknown and is estimated from the sample, while a z-test is used when the population standard deviation is known. The t-distribution is wider and has heavier tails than the z-distribution, which leads to a larger critical value and a higher chance of not rejecting the null hypothesis.</p> ####

3. Suppose we want to test the null hypothesis H0:p=0.20𝐻0:𝑝=0.20 against the alternative hypothesis Ha:p<0.20𝐻𝑎:𝑝<0.20 . Suppose also that we observed 25 successes in a random sample of 100 subjects and the level of significance is 0.05. What is the observed test statistic for this test?Multiple choice 1 Question 3  z=1.15𝑧=1.15   z=−1.25𝑧=−1.25   z=−1.15𝑧=−1.15   z=1.25

A sample of 39 observations is selected from a normal population. The sample mean is 43, and the population standard deviation is 6. Conduct the following test of hypothesis using the 0.02 significance level.

Suppose that the following is to be tested: H0: p = 0.72 and Ha: p ≠ 0.72. Calculate the test statistic for thefollowing sample data: Sixty-eight out of ninety test subjects have the characteristic of interest. Round tothe nearest thousandth