A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.39. A dietician is asked to test this claim and finds that a random sample of 24 servings has a variance of 1.47. Ata= 0.10, is there enough evidence to reject the manufacturer's claim? Assume the population is normally distributed. Complete parts (a) through (e) below.(a) Write the claim mathematically and identify H, and HOA. Ho: o? = 1.39 (Claim)H3i o? #1.39O B. Ho: o? #1.39H: o? = 1.39 (Claim)OC. Ho: o?s1.39 (Claim)H: o?> 1.39O D. H: o?21.39H: o? <1.39 (Claim)(b) Find the critical value(s) and identify the rejection region(s).The critical value(s) is(are)(Round to two decimal places as needed. Use a comma to separate answersneeded.)Choose the correct statement below and fill in the corresponding answer boxes.O A. The rejection regions are y2 < and y2 >O B. The rejection region is x? <.OC. The rejection region is x? >.(c) Find the standardized test statistic y(Round to two decimal places as needed.)(d) Decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Fill in the correct answers below.V Ho. ThereV enough evidence to at the 10% level of significance to reject the manufacturer's claim that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.39.Click to select your answer(s).MacBook Proesc@&1234678deleteQWEYUP}%3DASDFGHJKR

Name: A snack food manufacturer estimates that the variance of the number o
Uploaded: 2024-03-20T07:07:12.000Z
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Description: A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.39. A dietician is asked to test this claim and finds that a random sample of 24 servings has a variance of 1.47. At

In this context, the aim is to test whether the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.39. From the given information, the following values are known: a) The test hypothesis are: Here, the hypothesized variance is 1.39. The correct answer is, “Option A. H0: σ2=1.39(claim) Ha: σ2≠1.39” Sample size,n=24, Sample variance, s2=1.47, and Level of significance, α=0.10 b) In this context, the alternative is two-tailed. Therefore, the critical-value for the two-tailed test will be obtained as given below: Using EXCEL function, “=CHIINV(0.05,23)”, the critical value for χ20.05 is 35.172. Using EXCEL function, “=CHIINV(0.95,23)”, the critical value for χ20.95 is 13.091. The critical values are 13.091 and 35.172. The correct answer is “Option A, The rejection region are χ2<13.091 and χ2>35.172”.c) Test statistic: Test statistic=(n-1)s2σ2=(24-1)1.471.39=24.32 Since the test statistic is less than the critical value, the null hypothesis is not rejected. d) Therefore, the sample provides enough evidence to suggest that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.39. Fail to reject H0, there is enough evidence to at the 10% level of significance to reject the manufacturer’s claim that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.39.