According to the Pew Internet &American Life Project. 75% of American adults use the internet. The Pew project authors also peported on the percentage of Americans who use the internet by age group. The data in the in the agefile group was consistent with their findings. these data were obtained from a sample of 100 internet users in the 30-49 age group and 200 internet users in the 50-64 age group. A yes indicates the survey respondent had used the internet; a no indicates the survey respondent had not. formulate hypotheses that could be used to determine whether the percentage of internet users in the two age groups differs from the overall average of 75%.
Null hypothesis:
Ho: µ = .75 -->meaning population proportion equals .75 (or 75%)
Alternate or alternative hypothesis:
Ha: µ ≠ .75 -->meaning population proportion does not equal .75 (or 75%)
z=-2.26 => p = 0.0238
To determine whether the percentage of internet users in the two age groups differs from the overall average of 75%, we can formulate the following hypotheses:
Null Hypothesis (H0): The percentage of internet users in the 30-49 age group and 50-64 age group is equal to the overall average of 75%.
Alternative Hypothesis (Ha): The percentage of internet users in the 30-49 age group and 50-64 age group is different from the overall average of 75%.
This can be written as:
H0: p1 = p2 = 0.75
Ha: p1 ≠ p2 ≠ 0.75
where p1 represents the percentage of internet users in the 30-49 age group, p2 represents the percentage of internet users in the 50-64 age group, and 0.75 represents the overall average of internet users in American adults.
To formulate hypotheses that can determine whether the percentage of internet users in the two age groups differs from the overall average of 75%, we can use the following steps:
1. State the null hypothesis (H0): There is no difference between the percentage of internet users in the two age groups (30-49 and 50-64) and the overall average of 75%.
2. State the alternative hypothesis (Ha): The percentage of internet users in either the 30-49 age group or the 50-64 age group differs from the overall average of 75%.
Next, we need to determine the statistical test to be used. Since we are comparing two groups (30-49 and 50-64 age groups) to the overall average, we can use a two-sample proportion test.
3. Calculate the proportion of internet users in each age group:
- Proportion for the 30-49 age group (p1): Divide the number of internet users (100) by the total sample size (100) to get p1.
- Proportion for the 50-64 age group (p2): Divide the number of internet users (200) by the total sample size (200) to get p2.
4. Calculate the standard error of the difference between the proportions:
- Calculate the standard deviation for each group using the formula:
σ = √(p * (1 - p) / n), where p is the proportion and n is the sample size.
- Calculate the standard error for the difference between proportions using the formula:
SE = √((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))
5. Perform the two-sample proportion test:
- Calculate the test statistic, which follows a standard normal distribution:
z = (p1 - p2 - (p0 - p0)) / SE
p0 is the overall proportion (0.75), and p0 - p0 is 0 since we are comparing to the same value.
6. Determine the critical value or p-value:
- Depending on your chosen significance level (e.g., 0.05), compare the test statistic to the critical value from the standard normal distribution table or calculate the p-value associated with the test statistic.
7. Make a decision:
- If the test statistic falls within the rejection region (beyond the critical value) or the p-value is less than the chosen significance level, reject the null hypothesis.
- If the test statistic does not fall within the rejection region (not beyond the critical value) or the p-value is greater than the chosen significance level, fail to reject the null hypothesis.
By following these steps, you can formulate hypotheses and conduct a statistical test to determine whether the percentage of internet users in the 30-49 and 50-64 age groups differs from the overall average of 75%.