please help!!! thanks so much Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 23 cans of the soda drink. Those volumeshave a mean of 12.19 oz and a standard deviation of 0.11 oz, and they appear to be from a normally distributed population. If the workers want the fillingprocess to work so that almost all cans have volumes between 11.89 oz and 12.49 oz, the range rule of thumb can be used to estimate that the standarddeviation should be less than 0.15 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.15 oz. Usea 0.01 significance level. Complete parts (a) through (d) below.a. Identify the null and alternative hypotheses. Choose the correct answer below.O A. Ho: o = 0.15 ozВ. Но: о 3D0.15 ozH1: o<0.15 ozH1: 0+0.15 ozО С. Н: о20.15 ozH1: o<0.15 ozO D. Ho: o> 0.15 ozH1: 0 = 0.15 ozb. Compute the test statistic.x² =(Round to three decimal places as needed.)c. Find the P-value.P-value =(Round to four decimal places as needed.)d. State the conclusion.Ho, because the P-value isthe level of significance. There isevidence to conclude that the populationstandard deviation of can volumes is less than 0.15 oz.

Name: please help!!! thanks so much Workers at a certain soda drink factory
Uploaded: 2024-03-20T07:06:27.000Z
Channel: Bartleby
Description: please help!!! thanks so much Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 23 cans of the soda drink. Those volumeshave a mean of 12.19 oz and a standard deviation of 0.11 oz, and they

As due to the honor-code, we are authorized to answer only 3 subparts for each question and that you haven't specified which of the 3 subparts you want the solutions for, the first 3 subparts are being answered for now. Requesting you to re-upload the question stating the remaining subparts for the answers to them with all the required information. The claim is that the population of volumes has a standard deviation of less than 0.15 oz at 1% level of significance. Form the sample, standard deviation : s = 0.11 oz The null and alternative hypothesises are given as: H0: σ = 0.15 oz vs H1: σ < 0.15 oz The value of the test statistic is obtained as: ChiSquare : X2 = (N-1)( sσ0)2 =22 ×(0.110.15)2 =11.83111111 Therefore, Chi-Square = 11.831At α=0.01, the P-value of the test is obtained as: P-value = P(Chi-Square < ChiSquare0bserved = 11.8311111) = 0.960772 Hence, the p-value is 0.960772