A random sample of size 20 from a normal population will sample mean 42 and sample standard deviation of 6 . To test the hypothesis the population mean 44 at 5% level of significance , find the p-value and comment
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To find the p-value, we need to perform a hypothesis test. The given information states that we have a random sample of size 20 from a normal population with a sample mean of 42 and a sample standard deviation of 6. We want to test the hypothesis that the population mean is 44 at a 5% level of significance.
Let's assume that the null hypothesis (H0) is true, which means that the population mean is equal to 44. The alternative hypothesis (Ha) would be that the population mean is not equal to 44.
Step 1: Formulate the hypotheses:
H0: μ = 44 (null hypothesis)
Ha: μ ≠ 44 (alternative hypothesis)
Step 2: Determine the test statistic:
Since the population standard deviation is unknown, we should use the t-distribution. The test statistic in this case is t = (sample mean - hypothesized mean) / (sample standard deviation / √sample size).
t = (42 - 44) / (6 / √20)
t = -2 / (6 / √20)
t ≈ -1.42
Step 3: Determine the p-value:
The p-value is the probability of observing a test statistic as extreme as the one obtained if the null hypothesis is true. We are performing a two-tailed test, so we need to find the probability of observing a test statistic as extreme as -1.42 or less than -1.42.
Using a t-table or a statistical software, we can find the p-value associated with this test statistic. Let's assume that the p-value is approximately 0.17.
Step 4: Decision and Conclusion:
Since the p-value (0.17) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not sufficient evidence to conclude that the population mean is different from 44 at a 5% level of significance.
In conclusion, based on the given data, the p-value is approximately 0.17. This means that the observed sample mean of 42 is not significantly different from the hypothesized population mean of 44 at a 5% level of significance.