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qwelly [4]
3 years ago
15

A potato chip company makes potato chips in two flavors, Regular and Salt & Vinegar. Riley is a production manager for the c

ompany who is trying to ensure that each bag contains about the same number of chips, regardless of flavor. He collects two random samples of 10 bags of chips of each flavor and counts the number of chips in each bag. Assume that the population variances of the number of chips per bag for both flavors are equal and that the number of chips per bag for both flavors are normally distributed. Let the Regular chips be the first sample, and let the Salt & Vinegar chips be the second sample. Riley conducts a two-mean hypothesis test at the 0.05 level of significance, to test if there is evidence that both flavors have the same number of chips in each bag. (a) H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test. (b) t≈1.44 , p-value is approximately 0.167 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply. Select all that apply:
Mathematics
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

(a) H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test.

Step-by-step explanation:

We formulate the

H0: μ1=μ2; null hypothesis that the two means are equal and alternate hypothesis that the two mean are not equal.

Ha:μ1≠μ2 Two tailed test

Test statistic used is

t= x1`-x2` / s√n as the variances are equal and sample size is same

T value for 9 degrees of freedom for two tailed test at α = 0.05 is 2.26

P- value for t test for 9 degrees of freedom is 0.125 from the table.

Hence only a is correct .

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\displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\\\\CI=\biggr(\frac{87}{249}-\frac{58}{113}\biggr)\pm 2.33\sqrt{\frac{\frac{87}{249}(1-\frac{87}{249})}{249}+\frac{\frac{58}{113}(1-\frac{58}{113})}{113}}\\\\CI=\{-0.2941,-0.0337\}

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