Answer :
Final answer:
The test statistic for the given sample is -0.795. To find the p-value for this test statistic in a one-tailed test, you would reference the standard normal distribution, approximately yielding 0.2133. However, this p-value is an approximation, with exact values usually determined through statistical software.
Explanation:
To calculate the test statistic for the given sample, we use the formula for a one-sample z-test since the population standard deviation (σ) is known:
z = (M - μ) / (σ / √n)
Given the data:
Sample size (n): 34
The test statistic is calculated as follows:
z = (59.1 - 61.5) / (17.6 / √34)
z = -2.4 / (17.6 / 5.831)
z = -2.4 / 3.0183
z = -0.795
To find the p-value, we look up the z value in the standard normal distribution table, which gives us the probability that a standard normal variable is less than z. For z = -0.795, the p-value is approximately 0.2133. Since we are dealing with a left-tailed test, this is the p-value we would report. However, it should be noted that tables typically provide values for positive z-scores, and one might need to utilize software or a calculator's function for more precise calculation, especially since the standard normal table usually does not provide p-values up to four decimal places.
The reported p-value is an approximation, given that for hypothesis testing, precise p-values are commonly generated by software or statistical calculators which take the negative z-score and output the corresponding p-value directly for one-tailed tests.
Using the z-distribution, it is found that the desired measures are given as follows:
- The test statistic is z = -0.795.
- The p-value is of 0.2133.
What are the hypotheses tested?
At the null hypotheses, it is tested if the mean is of 61.5, that is:
[tex]H_0: \mu = 61.5[/tex]
At the alternative hypotheses, it is tested if the mean is less than 61.5, that is:
[tex]H_1: \mu < 61.5[/tex]
What is the test statistic?
The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- [tex]\sigma[/tex] is the standard deviation of the population.
- n is the sample size.
The values of the parameters are given as follows:
[tex]\overline{x} = 59.1, \mu = 61.5, \sigma = 17.6, n = 34[/tex]
Hence, the test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{59.1 - 61.5}{\frac{17.6}{\sqrt{34}}}[/tex]
z = -0.795.
What is the p-value?
We have a left-tailed test, as we are testing if the mean is less than a value, with z = -0.795. Hence, using a z-distribution calculator, the p-value is of 0.2133.
More can be learned about the z-distribution at https://brainly.com/question/16162795
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