Problem

$\checkmark$ e human resources department of a large investment bank announced that the number of people it interviews monthly has a mean of 100 with a standard deviation of 16.5. The management of the bank suspects that the standard deviation exceeds 16.5 . Suppose that the management wants to take a small sample of months and carry out a hypothesis test to see if its suspicions hav support. State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ that it would use for this test. \[ \begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array} \] \begin{tabular}{ccc} $\mu$ & $\bar{x}$ & $\rho$ \\ $\hat{\rho}$ & $\sigma$ & $s$ \\ $\square^{p}$ & & $\square<\square$ \\ $\square \leq \square$ & $\square>\square$ & $\square \geq \square$ \\ $\Pi=\Pi$ & $\square \neq \Pi$ & \end{tabular} Explanation Check

Solution

Step 1 :In hypothesis testing, the null hypothesis (H0) is the statement that is assumed to be true unless there is strong enough evidence to reject it. The alternative hypothesis (H1) is the statement that we are testing for, it is the statement that we will accept if there is enough evidence to reject the null hypothesis. In this case, the management of the bank suspects that the standard deviation of the number of people it interviews monthly exceeds 16.5. Therefore, the null hypothesis (H0) would be that the standard deviation is equal to 16.5, and the alternative hypothesis (H1) would be that the standard deviation is greater than 16.5. So, the hypotheses would be stated as follows: H0: σ = 16.5 H1: σ > 16.5 Here, σ represents the standard deviation of the population. The null hypothesis states that the standard deviation is equal to 16.5, while the alternative hypothesis states that the standard deviation is greater than 16.5. These hypotheses will be tested using the sample data collected by the bank's management. If the sample data provides strong enough evidence to reject the null hypothesis, then the management can conclude that their suspicion is correct and the standard deviation does indeed exceed 16.5.

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Source: https://solvelyapp.com/problems/N45gHVuzaQ/

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