To test H0: mu = 100 versus Ha: mu 100, a simple random s...

Question

To test H0: mu = 100 versus Ha: mu > 100, a simple random sample of size n = 35 is obtained from an unknown distribution. The sample mean is 104.3 and the sample standard deviation is 12.4.

b. Use the classical or p-value approach to decide whether to reject the statement in the null hypothesis at the alpha = 0.05 level of significance.

Accepted Answer

Hello, everyone, let's take a look at this question together. A researcher claims that the mean annual salary for a certain profession is at least 60,000. At the 0.05 significance level, test this claim using the following sample statistics. Our sample mean is equal to 58,800. Our sample standard deviation is equal to 2400, and our sample size is equal to 16. Assume the population is normally distributed. What is the result of the hypothesis test? So in order to solve this question, we have to conduct a hypothesis test where the researcher claims that the mean annual salary for a certain profession is at least $$60,000 and at the 0.05 significance level, we're testing the claim using the sample statistics that are given, which is a sample mean of 58$$,800 a sample standard deviation of $$2400 and a sample. Size of 16 and based on the provided information, assuming that the sample is random since the population standard deviation is unknown and the population is normally distributed, we know that we can use the T test. And so the first step is stating the null and alternative hypotheses, which we know the researcher claims that the mean annual salary for a certain profession is at least. 60$$,000 which this can be expressed as mu is greater than or equal to $$60,000 and since it is a statement of strict equality, it is the null hypothesis. So the null hypothesis is mu is greater than or equal to 60,000, and we know the complement of the null hypothesis is the alternative hypothesis, so our alternative hypothesis is given as mu is less than. 60,000. Therefore, the null and alternative hypotheses are a null hypothesis of mu is greater than or equal to 60,000, which is the claim, and an alternative hypothesis, which is mu is less than 60,000. And from the information provided, the given level of significance, which is alpha, is alpha equals 0.05, and the next step is to identify the degrees of freedom. which we know that the degrees of freedom are calculated as N minus 1, so plugging in 16 for N, 16 minus 1 is equal to 15, so our degrees of freedom is 15. Then the next step is to determine the critical value which is represented as T sub-zero, and since our alternative hypothesis contains the less than inequality symbol, the hypothesis test is a left-tailed test. So we will use the table for T distribution and look for the intersection of our degrees of freedom, which equals 15, and our significance level alpha, which equals 0.05, since this is a one tail test, and from the distribution table, we can identify that our critical value of T. Subzero is equal to 1.753, and we should note that we need to negate the critical value, since this is a left tail test, so our critical value of T subzero is actually equal to -1.753. And now we can determine the rejection region. And since the test is a left-tailed test and the critical value is -1.753, the rejection region is T is less than -1.753, and our next step is to find the standardized test statistic, which is represented as the variable T and is calculated as T is equal to our sample mean subtracted by the population mean divided by the sample standard DV. which is divided by the square root of our sample size N, and substituting the values into the formula, we have 58,800 subtracted by 60,000, which is divided by 2400 divided by the square root of 16. And when simplified, our test statistic T is equal to -2. And so using both the critical value and the standardized test statistic, we can make our decision and since our Standardized test statistic T is less than the critical value in a left tailed test, T is in the rejection region. Therefore, we reject the null hypothesis. And since the claim is the null hypothesis, and we reject the null hypothesis, the interpretation for this decision is given as there is sufficient evidence to reject the claim that the mean annual salary for a certain profession is at least 60$$,000. So the result of the hypothesis test is that there is sufficient evidence to reject the claim that the mean annual salary for a certain profession is at least $60,000 which looking at our answer choices, we can identify that this matches answer choice A, so answer choice A is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Hypothesis Testing

t-Test for a Single Mean

p-Value and Significance Level

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