Can you help me answer the question by using my material that shows on pictures? Can you show the calculation for finding the answer? Material = 145-1 until 145-8

FF ' f 145] oma's pretzels revisited In the following exercise you will revisit an earlier problem, this time solving it with the p~value method. 1452 oma's pretzels revisited Blanche McCarthy is the marketinc director of Oma's Own snack food company. Oma's makes toasted pretzel snacks. Each bag of pretzels contains one serving, and Oma's advertises that the pretzel snacks contain an average of 112 calories per serving, 145-3 oma's pretzels revisited In a recent test, an independent consumer research organization conducted an experiment to see if this claim was true. The researchers found that the average calorie content in a sample of 32 bags was 102 calories per sewing. The standard deviation of the sample was 1 g. 145-4 oma's pretzels revisited Blanche would like to know if the calorie content of Oma's pretzels has really changed. so she can market them appropriately. At the signicance level 0.01, do \"up these data indicate that the pretzels' calorie content has changed? (75 A -.No. 0-. @Yes. C.@Can't be determined otn the information given. 1 1 anl ? ? 32 Sample Mean (2) 102ml Sample Std. Dev. (5) 19cal Significance Level 0.01 145-5 oma's pretzels revisited In this problem, the null hypothesis is that the actual population mean is what Oma's has always advertised. A two-side test is more appropriate in this problem, since Blanche only wants to know if the mean calorie content has changed. 145-4 oma's pretzels revisited Blanche would like to know if the calorie content of Oma's pretzels has really changed, so she can market them appropriately. At the significance level 0.01, do nes tian these data indicate that the pretzels' calorie content has changed? A.ONo. B.OYes. C-Can't be determined from the information given. Null Hypothesis Mean 112cal Null Hypothesis ? Alternative Hypothesis ? Sample Size (n) 32 Sample Mean (X) 102cal Sample Std. Dev. (s) 19cal Significance Level 0.01 145-5 oma's pretzels revisited In this problem, the null hypothesis is that the actual population mean is what Oma's has always advertised. A two-side test is more appropriate in this problem, since Blanche only wants to know if the mean calorie content has changed. Null Hypothesis Mean 112cal Null Hypothesis 1 = 112cal Alternative Hypothesis H # 112cal Sample Size (n) 32 Sample Mean (x) 102cal Sample Std. Dev. (s) 19cal Significance Level 0.01 145-6 oma's pretzels revisited Assuming that the null hypothesis is true, you find a z-value for the sample mean of 102 using the appropriate formula. The z-value is -2.98. Null Hypothesis Mean 112cal Null Hypothesis H = 112cal Alternative Hypothesis U = 112cal Sample Size (n) 32 Sample Mean (x) 102cal Sample Std. Dev. (s) 19ca Significance Level 0.01 X - ul 102 - 112 Z = = -2.98 S 19 132145-6 oma's pretzels revisited Assuming that the null hypothesis is true, you find a z-value for the sample mean of 102 using the appropriate formula. The z-value is -2.98. Null Hypothesis Mean 112cal Null Hypothesis H = 112cal Alternative Hypothesis p # 112cal Sample Size (n) 32 Sample Mean (x) 102cal Sample Std. Dev. (s) 19cal Significance Level 0.01 X - ul 102 - 112 Z = = -2.98 S 19 V32 145-7 oma's pretzels revisited Using the Excel NORMSDIST function or the Standard Normal Table, you can find the corresponding left-tail probability of 0.00 14. For a two-sided test, you double this number to find the p-value, in this case 0.0028, Null Hypothesis Mean 112cal Null Hypothesis H = 112cal Alternative Hypothesis H # 112cal Sample Size (n) 32 Sample Mean (x) 102cal Sample Std. Dev. (s) 19cal Significance Level 0.01 0.0014 0.0014 -2.98 2.98 145-8 oma's pretzels revisited Since this p-value is less than the significance level, you can reject the null hypothesis, Moreover, you now can say that you are rejecting the null hypothesis at the 0.0028 level of significance. You can recommend to Blanche that she have the labeling changed on the pretzel bags, and adjust her marketing accordingly. 0.0028