A research article reported that for a random sample of 850 meal purchases made at fast food chain A, the mean sodium content was 1,684 mg, and the standard deviation was 824 mg. For a random sample of 2,109 meal purchases made at fast food chain B, the mean sodium content was 1,479 mg, and the standard deviation was 817 mg. Based on these data, is it reasonable to conclude that there is a difference in mean sodium content for meal purchases at fast food chain A and meal purchases at fast food chain B? Use 𝛼 = 0.05. (Use 𝜇1 for fast food chain A and 𝜇2 for fast food chain B.)
a) State the appropriate null and alternative hypotheses (in mg).
b) Find the test statistic. (Round your answer to two decimal places.)
c) Use technology to find the P-value. (Round your answer to four decimal places.)
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a) The appropriate null and alternative hypotheses are:
Null hypothesis (H0): 𝜇1 = 𝜇2 (There is no difference in mean sodium content for meal purchases at fast food chain A and B.)
Alternative hypothesis (H1): 𝜇1 ≠ 𝜇2 (There is a difference in mean sodium content for meal purchases at fast food chain A and B.)
b) The test statistic can be calculated using the formula:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Plugging in the given values, we get:
t = (1684 - 1479) / sqrt((824^2 / 850) + (817^2 / 2109))
≈ 4.52
c) Using technology, we can find the p-value associated with the test statistic. The p-value is the probability of obtaining a test statistic as extreme as the calculated test statistic under the null hypothesis.
By using a t-distribution calculator or software, we can find that the p-value is approximately 0.0000 (rounded to four decimal places).
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