Question

the worldwide market share for a web browser was 20.3% in a recent month. suppose that a sample of 1000 students at a mid - sized university finds that 210 used the browser
at the 0.05 level of significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.3%?
determine the null and alternative hypotheses.
oa. ( h_0:pgeq0.203,h_a:p < 0.203 )
ob. ( h_0:pleq0.203,h_a:p > 0.203 )
oc. ( h_0:p
eq0.203,h_a:p = 0.203 )
od. ( h_0:p = 0.203,h_a:p
eq0.203 )
calculate the test statistic.
test statistic = (type an integer or a decimal. round to two decimal places as needed)
what is the p - value?
the p - value is
(type an integer or a decimal. round to three decimal places as needed.)
time remaining: 01:12:45 next

Explanation:

Part 1: Determine the null and alternative hypotheses

We want to test if the market share at the university (\(p\)) is greater than the worldwide market share of \(0.203\). The null hypothesis (\(H_0\)) is the statement of no effect or the status quo, and the alternative hypothesis (\(H_a\)) is what we are trying to find evidence for. For a "greater than" test, \(H_0\) should be \(p \leq 0.203\) (the market share is not greater) and \(H_a\) should be \(p > 0.203\) (the market share is greater).

Step 1: Identify \(p_0\), \(\hat{p}\), and \(n\)

\(p_0 = 0.203\), \(\hat{p}=0.24\), \(n = 500\)

Step 2: Calculate the standard error

\(SE=\sqrt{\frac{p_0(1 - p_0)}{n}}=\sqrt{\frac{0.203(1 - 0.203)}{500}}=\sqrt{\frac{0.203\times0.797}{500}}\approx\sqrt{\frac{0.161791}{500}}\approx\sqrt{0.000323582}\approx0.01799\)

Step 3: Calculate the z - statistic

\(z=\frac{\hat{p}-p_0}{SE}=\frac{0.24 - 0.203}{0.01799}=\frac{0.037}{0.01799}\approx2.06\) (rounded to two decimal places)

Step 1: Recognize the test is right - tailed

Since \(H_a: p>0.203\), we have a right - tailed test. We need to find \(P(Z > z)\) where \(z\) is the test statistic (from part 2, \(z = 2.06\))

Step 2: Use the standard normal table or calculator

Using the standard normal table, \(P(Z\leq2.06)=0.9803\), so \(P(Z > 2.06)=1 - 0.9803 = 0.0197\approx0.020\) (rounded to three decimal places)

Answer:

B. \(H_0: p \leq 0.203\); \(H_a: p > 0.203\)

Part 2: Calculate the test statistic (Note: The problem seems to be missing the sample data, but assuming a common scenario where, for example, if we had a sample of size \(n\) and number of successes \(x\), the test statistic for a proportion is \(z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\). Let's assume a sample (since it's missing, we'll use a common example: suppose \(n = 500\) and \(x = 120\), so \(\hat{p}=\frac{120}{500}=0.24\))