A student weighs 10 candy bars; test whether the average weight differs from the expected value using a t-test. Which test is appropriate?

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Multiple Choice

A student weighs 10 candy bars; test whether the average weight differs from the expected value using a t-test. Which test is appropriate?

Explanation:
When you want to know if a sample mean differs from a known value, you use a one-sample t-test. Here you have a single sample of ten candy bars and you’re comparing the average weight to the expected mean, with the population standard deviation unknown. With n = 10, the degrees of freedom are n − 1, which is 9, so the test uses nine degrees of freedom. The test statistic is t = (x̄ − μ0) / (s/√n), reflecting how far the observed mean is from the expected mean in units of the standard error. If the population standard deviation were known, a Z-test would be used, but that’s rare in practice. It’s not a two-sample test since there aren’t two independent groups to compare, and it’s not a paired test because there’s no pairing of measurements. So the appropriate method is a one-sample t-test with nine degrees of freedom.

When you want to know if a sample mean differs from a known value, you use a one-sample t-test. Here you have a single sample of ten candy bars and you’re comparing the average weight to the expected mean, with the population standard deviation unknown. With n = 10, the degrees of freedom are n − 1, which is 9, so the test uses nine degrees of freedom. The test statistic is t = (x̄ − μ0) / (s/√n), reflecting how far the observed mean is from the expected mean in units of the standard error. If the population standard deviation were known, a Z-test would be used, but that’s rare in practice. It’s not a two-sample test since there aren’t two independent groups to compare, and it’s not a paired test because there’s no pairing of measurements. So the appropriate method is a one-sample t-test with nine degrees of freedom.

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