# What is the test statistic and the p-value for the following sample: H0: μ = 60 versus H1: μ _= 60, α = .025, ¯x = 63, σ = 8, n = 16?

test statistic = 1.5

p-value = 0.1336

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To find the test statistic and p-value for the given sample with hypothesis testing, we can use the z-test because the population standard deviation (σ) is known, and the sample size (n) is greater than 30.

First, we calculate the test statistic (z-score) using the formula:

[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} ]

where:

- ( \bar{x} ) is the sample mean,
- ( \mu ) is the population mean under the null hypothesis,
- ( \sigma ) is the population standard deviation,
- ( n ) is the sample size.

Given:

- ( \bar{x} = 63 ),
- ( \mu = 60 ),
- ( \sigma = 8 ),
- ( n = 16 ).

Substituting the values:

[ z = \frac{63 - 60}{\frac{8}{\sqrt{16}}} = \frac{3}{2} = 1.5 ]

Next, we find the p-value associated with this z-score using a standard normal distribution table or statistical software. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from the sample, assuming the null hypothesis is true.

For a two-tailed test with α = 0.025, the critical z-value is approximately ±1.96.

Since the calculated z-score of 1.5 falls within the range of -1.96 to 1.96, we fail to reject the null hypothesis.

The p-value for a z-score of 1.5 is the area under the standard normal distribution curve to the right of 1.5 (since it's a one-tailed test). Using a standard normal distribution table or statistical software, we find the corresponding p-value, which represents the probability of observing a value as extreme as 1.5 or greater under the null hypothesis.

The calculated p-value will determine whether to reject or fail to reject the null hypothesis. If the p-value is less than or equal to the significance level (α), we reject the null hypothesis; otherwise, we fail to reject it.

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