In 2006, 75.9% of first-year students said they used the Internet for research or homework. Administrators found that 168 of an SRS of 200 first-year students used the Internet for research or homework. Is the proportion of first-year students who used the Internet for research or homework larger than the 2006 national value of 75.9%?
Also, carry out the hypothesis test described above and compute the P-value, and how does your value compare with the value given above?
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 = .759 -->meaning: population proportion is equal to .759 (converting the 75.9% to a decimal).
Alternative hypothesis:
Ha: p > .759 -->meaning: population proportion is greater than .759 (this is a one-tailed test).
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .84 - .759 -->test value (168/200 is .84) minus population value (.759)
divided by
√[(.759)(.241)/200] --> .241 represents 1-.759 and 200 is sample size.
Finish the calculation. Remember if the null is not rejected, then there is no difference. If you need to find the p-value for the test statistic, check a z-table. The p-value is the actual level of the test statistic.
I hope this will help get you started.
p=0.0031
To determine if the proportion of first-year students who used the Internet for research or homework is larger than the 2006 national value of 75.9%, we can conduct a hypothesis test.
Step 1: State the hypotheses.
- Null hypothesis (H0): The proportion of first-year students who used the Internet for research or homework is equal to 75.9%.
- Alternative hypothesis (Ha): The proportion of first-year students who used the Internet for research or homework is larger than 75.9%.
Step 2: Set the significance level.
Let's assume a significance level of α = 0.05.
Step 3: Calculate the test statistic.
We will use a one-sample proportion z-test, where the test statistic is calculated as:
z = (p - p0) / sqrt(p0*(1-p0)/n)
where:
- p is the sample proportion of first-year students who used the Internet for research or homework,
- p0 is the hypothesized proportion (75.9%),
- n is the sample size.
In this case, p = 168/200 = 0.84, p0 = 0.759, and n = 200.
Step 4: Calculate the p-value.
The p-value is the probability of observing a sample proportion as extreme as (or more extreme than) the one obtained, assuming the null hypothesis is true. Since we are conducting a one-tailed test to determine if the proportion is larger, we will calculate the upper-tail p-value.
To calculate the p-value, we can use a standard normal distribution table or statistical software. The z-value can be computed as:
z = (0.84 - 0.759) / sqrt(0.759*(1-0.759)/200) = 5.215
Using a standard normal distribution table or software, we find that the p-value corresponding to a z-value of 5.215 is very close to 0.
Step 5: Make a decision.
Compare the p-value with the significance level. Since the p-value (close to 0) is smaller than the significance level (0.05), we reject the null hypothesis.
Step 6: Conclusion.
We have sufficient evidence to suggest that the proportion of first-year students who used the Internet for research or homework is larger than the 2006 national value of 75.9%. The computed p-value is close to 0, indicating a highly significant result.
To determine if the proportion of first-year students who used the Internet for research or homework is larger than the 2006 national value of 75.9%, we need to carry out a hypothesis test.
Step 1: State the hypotheses
The null hypothesis (H₀): p = 0.759 (proportion of first-year students using the Internet for research or homework is equal to 75.9%)
The alternative hypothesis (H₁): p > 0.759 (proportion of first-year students using the Internet for research or homework is larger than 75.9%)
Step 2: Set the significance level
Let's assume a significance level (α) of 0.05.
Step 3: Conduct the test and calculate the test statistic
We have a random sample of 200 first-year students, where 168 of them use the Internet for research or homework. The sample proportion (p̂) is calculated as 168/200 = 0.84.
Using the formula for the test statistic for a proportion:
z = (p̂ - p₀) / √((p₀(1-p₀)) / n)
where p̂ is the sample proportion, p₀ is the hypothesized proportion, and n is the sample size.
Plugging in the values:
z = (0.84 - 0.759) / √((0.759(1-0.759)) / 200)
Calculating this expression will give us the test statistic (z-value).
Step 4: Calculate the p-value
The p-value is the probability of observing a test statistic as extreme as the one computed (or more extreme) under the null hypothesis.
Since we are testing whether the proportion is larger, we will calculate the p-value as the probability of observing a value greater than the test statistic. We can use a standard normal distribution table or a calculator to find this probability (area under the curve to the right of the test statistic).
Step 5: Make a decision and interpret the results
If the p-value is less than the significance level (α), we reject the null hypothesis in favor of the alternative hypothesis. If the p-value is greater than α, we fail to reject the null hypothesis.
In this case, comparing the calculated p-value with the significance level (α = 0.05), we can make a decision.
Now, let's perform the calculations to determine the test statistic (z-value), calculate the p-value, and interpret the results.