**Hypothesis Testing for Mean Customer Satisfaction****Introduction:**Over the years, the mean customer satisfaction rating at a local restaurant has been 92. The restaurant was recently remodeled, and the management claims the mean customer rating, μ, is not equal to 92. In a sample of 68 customers chosen at random, the mean customer rating is 94.3. Assume that the population standard deviation of customer ratings is 14.3.**Objective:**Is there enough evidence to support the claim that the mean customer rating is different from 92? Perform a hypothesis test using the 0.05 level of significance.**Procedure:**(a) **State the Hypotheses:**- Null hypothesis (H₀): μ = 92- Alternative hypothesis (H₁): μ ≠ 92(c) **Perform a Hypothesis Test:**- The test statistic has a normal distribution (Z-test).- z₀.₀₂₅ is the value that cuts off an area of 0.025 in the right tail.- The test statistic is calculated using the formula: \ \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]**Graph Explanation:**- The diagram presents a standard normal distribution curve.- It shows areas of significance typically in two tails, indicating the rejection regions for a two-tailed test at the 0.05 level of significance.**Step-by-step Process:**1. **Select the type of test:** - One-tailed - Two-tailed (shown in use here)2. **Place significant boundaries:** The critical values (z-scores) determine the regions where we reject the null hypothesis if the test statistic falls within them.**Conclusion:**Assess whether the calculated z-value falls within the critical regions as per the standard normal distribution. If it does, there is adequate evidence to reject the null hypothesis and accept the alternative hypothesis, concluding that the mean customer rating is indeed different from 92. **Hypothesis Testing with the Standard Normal Distribution****Instructions:**1. **Formulate a Hypothesis Test:** - The test statistic follows a normal distribution, making it suitable for a "Z-test." - \( z_{0.025} \) is the value that cuts off an area of 0.025 in the right tail. - The test statistic is calculated using: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]2. **Standard Normal Distribution Process:** - **Step 1:** Select either a one-tailed or two-tailed test. - Options: One-tailed or Two-tailed (radio buttons available for selection) - **Step 2:** Enter the critical value(s), rounded to three decimal places. - **Step 3:** Enter the test statistic, rounded to three decimal places.**Graph Explanation:**The graph displayed is a standard normal distribution curve. The x-axis is labeled with values ranging from -3 to 3, and the y-axis represents the probability density. The curve peaks at 0, which is the mean of the distribution, and symmetrically tails off as you move towards -3 and 3.**Conclusion:**- **Step for Conclusion:** - Based on your answer to part (b), determine the conclusion at a 0.05 level of significance regarding the management's claim. - If the value of the test statistic falls in the rejection region, the null hypothesis is rejected. Consequently, there is sufficient evidence to support the claim.**Options:**- Buttons for "Explanation" and "Check" can be used to further understand or verify the calculations.---This section is part of a statistical educational module intended to help students understand the process of conducting hypothesis tests using the standard normal distribution.

Parametric test is used when the population follows the normal distribution. Some of the parametric test are z test, t test, chi square test, F test etc. If the population variance are known to test mean significance, then z test is used. If the population variance are unknown to test mean significance, then t test is used. a) Given : Mean customer satisfaction rating is 92 Sample of 68 customers are taken and mean customer rating is 94.3 Population standard deviation of customer rating is 14.3 Claim : Mean customer rating is different from 92 Let the null and alternate hypothesis : H0 : μ=92H1 : μ≠92b) Given : Step 1 : Since the claim represent the different value. I.e. Two tailed used Option B is correct Incorrect option : Option A - Less than or greater than symbol not used Step 2 : z0.025 of right tail is 1.960 (From the normal distribution table) Step 3 : Test statistic : z=x¯-μσn=94.3-9214.368=1.326 Test statistic is 1.326 Here, the shaded region denote the rejection region.c) From the part (b), Standard normal distribution plot, The value of test statistic does not lie in the rejection region. Thus, the null hypothesis failed to be rejected. There is not enough evidence to conclude that the mean customer rating is different from 92.