a.) Using technology (calculator or StatCrunch), find the sample mean and sample standard deviation. (Round two decimal places).b.) H0: __________________ H1: __________________Type of Test: ________________Test Statistic: _________________ (2 decimal places)Critical Value(s): ________________ (3 decimal places)P-Value: __________________ (4 decimal places)Label the critical value and test statistic on the curve and shade the rejection region(s).Reject or Fail to Reject: ______________________Conclusion (express in a sentence in the context of the problem) Many consumers pay careful attendant to stated nutritional content on packaged foods when makingpurchases, so it is important that the information on packages be accurate. A random sample of 14 frozendinners of a certain type was selected and the calorie content of each one was determined. Below are theresulting observations. Assume the data is normally distributed. The stated mean calorie content is 240.Perform a hypothesis test to determine if the actual mean calorie content is different from the stated packagevalue of 240. Use a 0.01 level of significance.255 244 239 242 265 245 243259 248 225226 254 233 235

Name: a.) Using technology (calculator or StatCrunch), find the sample mean
Uploaded: 2024-03-20T07:06:16.000Z
Channel: Bartleby
Description: a.) Using technology (calculator or StatCrunch), find the sample mean and sample standard deviation. (Round two decimal places).b.) H0: __________________ H1: __________________Type of Test: ________________Test Statistic: _________________ (2 decima

a) For the given data mean and standard deviation is calculated using Ti 83+ is as follows: Choose Stat>Edit>Edit. Enter the given values in List L1. Select Stat>Calc>1 Var stats. Select ‘2nd’ and then press 1. Output: Sample Mean is 243.79 Sample standard deviation is 11.83b) Hypothesis: H0:μ=240H1:μ≠240 Two tailed test. Since population standard deviation is not known. it is appropriate to use One sample t test. Procedure for 1 sample t test using TI-83+ calculator: Select STAT>TESTS>T test Select Input as DATA Enter µ0 as 240 Select List as L1 and select µ≠ µ0 Select calculate. Output screen: Thus, Test statistic is 1.19 p-value is 0.2527 Critical values: Computation of critical-value: The two tailed critical-value for the t-test at 13 degrees of freedom and 0.01 level of significance is 3.012 can be obtained using the excel formula, “=T.INV.2T(0.01,13)” Critical values are ±3.012.Graph: Decision rule: If p-value ≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: Here, p-value (0.2527) is greater than the level of significance (0.01). Therefore, fail to reject the null hypothesis. The actual mean calorie content is same from the stated package value of 240.