Over the years, the mean customer satisfaction rating at a local restaurant has been 92. The restaurant was recently remodeled, and now the management claims the mean customer rating, \( \mu \), is not equal to 92. In a sample of 68 customers chosen at random, the mean customer rating is 88.2. Assume that the population standard deviation of customer ratings is 14.3.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.### (a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \).- **\( H_0 \)**: \( \mu = 92 \)- **\( H_1 \)**: \( \mu \neq 92 \)### (b) Perform a Z-test and find the p-value.Here is some information to help you with your Z-test.- The value of the test statistic is given by: \[\frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]- The p-value is two times the area under the curve to the left of the value of the test statistic.### Diagram ExplanationThe diagram shows a standard normal distribution curve with steps to find the p-value:1. Select one-tailed or two-tailed.2. Enter the test statistic (round to 3 decimal places).3. Shade the area represented by the p-value.4. Enter the p-value (round to 3 decimal places).### (c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by the management.- If the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. There is enough evidence to support the claim that the mean customer rating is not equal to 92.- If the p-value is greater than the level of significance, the null hypothesis is not rejected. There is not enough evidence to support the claim that the mean customer rating is not equal to 92. Please choose the correct conclusion based on your calculated p-value. According to a report done by S & J Power, the mean lifetime of the light bulbs it manufactures is 42 months. A researcher for a consumer advocate group tests this by selecting 32 bulbs at random. For the bulbs in the sample, the mean lifetime is 41 months. It is known that the population standard deviation of the lifetimes is 8 months. Assume that the population is normally distributed. Can we conclude, at the 0.10 level of significance, that the population mean lifetime, μ, of light bulbs made by this manufacturer differs from 42 months?Perform a two-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.)(a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \).\[ H_0 : \mu = 42 \]\[ H_1 : \mu \neq 42 \](b) Determine the type of test statistic to use.[Dropdown: Z](c) Find the value of the test statistic. (Round to three or more decimal places.)[Input box for calculation](d) Find the two critical values. (Round to three or more decimal places.)\[ -1.645 \text{ and } 1.645 \](e) Can we conclude that the population mean lifetime of light bulbs made by this manufacturer differs from 42 months?[Option buttons: Yes No (selected)]

"Since you have posted multiple questions, we will provide the solution only to the first question as per our Q&A guidelines. Please repost the remaining questions separately.”The sample size is, n =68The mean for the randomly selected samples is, X¯ = 88.2The population standard deviation is, σ=14.3The defined level of significance is, α=0.05It is required to determine whether the mean customer rating is different from 92a) Since it is to check whether the sample mean is different from 92 or not , it is suitable to consider the two-tailed hypothesis and this can be written asNull Hypothesis, H0:μ=92 : The mean customer rating is equal to 92.Alternative Hypothesis, Ha:μ≠92 : The mean customer rating is not equal to 92.b) Becasue the sample size is larger (n>30) and population variance is known, the above hypothesis can tested using the one-sample Z-test and the test statistic is given byThe p-value can be determined as folllowsthis probability can be determined using the EXCEL function =2*NORM.S.DIST(-2.1913,1)Therefore the p-value is 0.02843The plot of p-value on the normal curve is as followsc) Rejection region based on the p-valueIf p-value < 0.05, then reject the null hypothesisIf p-value ≥ 0.05, then conclude that no strong evidence to reject null hypothesisConclusion:Since the p-value, 0.02843 < 0.05, reject null hypothesis and conclude that there is enough evidence to support that the mean ratings is not equal to 92Hence the correct option is (A) Since the p-valueis less than level of significance, the null hypothesis is rejected, there is enough evidence to support the claim that the mean ratings is not equal to 92a) The null and alternative hypothesis areH0:μ=92H0:μ≠92b) The Z- test statistic resultsstep 1 : two -tailedstep 2: The vaule of the test statistic is -2.191step 3: the shaded region of the p-value is step 4: The p-value is 0.028c) (A) Since the p-valueis less than level of significcance, the null hypothesis is rejected, there is enough evidence to support the claim that the mean ratings is not equal to 92

Math

Statistics

Over the years, the mean customer satisfaction rating at a local restaurant has been 92. The restaurant was recently remodeled, and now 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 88.2. Assume that the population standard deviation of customer ratings is 14.3. 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. (a) State the null hypothesis Ho and the alternative hypothesis H₁. Ho: H₁:0 (b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. • The value of the test stati Standard Normal Distribution Step 1: Select one-tailed or two-tailed. O One-tailed O Two-tailed Step 2: Enter the test statistic. (Round to 3 decimal places.) Step 3: Shade the area represented by the p-value. given by Step 4: Enter the p-value. (Round to 3 decimal places.) x-μ 0 μ □<口 020 X • The p-value is two times the area under the curve to the left of the value of the test statistic. X OSO 0=0 ☐#0 0.3+ O>O 0.2+ 0.1+ x (c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by the management. O Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean customer rating is not equal to 92. O Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean customer rating is not equal to 92. O Since the p-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean customer rating is not equal to 92. O Since the p-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean customer rating is not equal to 92.