From a random sample of 500 males interviewed, 125 indicated that they watch professional football on Monday night television. Does this evidence indicate that more than 20% of the male TV viewers watch professional football on Monday evenings? Use the 0.01 level of significance, and find the p-value.
You can try a proportional one-sample z-test for this one since this problem is using proportions.
Here's a few hints to get you started:
Null hypothesis:
Ho: p = .20 -->meaning: population proportion is equal to .20
Alternative hypothesis:
Ha: p > .20 -->meaning: population proportion is greater than .20
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .25 - .20 -->test value (125/500 is .25) minus population value (.20) divided by
√[(.25)(.75)/500] --> .75 represents 1-.25 and 500 is sample size.
Finish the calculation. Remember if the null is not rejected, then you cannot conclude p > .20. If you need to find the p-value for the test statistic, check a z-table, then compare to .01 to determine whether or not to reject the null.
I hope this helps.
This section should read:
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .25 - .20 -->test value (125/500 is .25) minus population value (.20) divided by
√[(.20)(.80)/500] --> .80 represents 1-.20 and 500 is sample size.
Sorry for any confusion.
To determine whether the evidence indicates that more than 20% of male TV viewers watch professional football on Monday evenings, we can conduct a hypothesis test. Here are the steps to follow:
Step 1: Formulate the hypotheses:
- Null Hypothesis (H0): The proportion of male TV viewers who watch professional football on Monday evenings is equal to or less than 20% (p ≤ 0.20).
- Alternative Hypothesis (Ha): The proportion of male TV viewers who watch professional football on Monday evenings is greater than 20% (p > 0.20).
Step 2: Determine the test statistic:
We will use the Z-test for proportions since we have a sample proportion and want to compare it to a hypothesized proportion.
Step 3: Calculate the test statistic:
The formula for the test statistic (Z-score) is:
Z = (p̂ - p0) / √(p0(1-p0) / n)
where:
- p̂ is the observed proportion (125/500 = 0.25)
- p0 is the hypothesized proportion (0.20)
- n is the sample size (500)
Calculating the Z-score:
Z = (0.25 - 0.20) / √(0.20(1-0.20) / 500)
Z ≈ 1.77
Step 4: Determine the p-value:
To find the p-value, we need to calculate the area under the normal distribution curve to the right of the calculated Z-score (1.77).
Using a Z-table or a statistical software, we find that the p-value associated with a Z-score of 1.77 is approximately 0.038.
Step 5: Make a decision:
Using a significance level (α) of 0.01, if the p-value is less than α, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since the p-value (0.038) is greater than the significance level (0.01), we fail to reject the null hypothesis. Therefore, the evidence does not indicate that more than 20% of male TV viewers watch professional football on Monday evenings.
The p-value is approximately 0.038.