Question

A random sample of 150 women shows that 45 are against certain new working hours. A newspaper claims that less than 33% of women are against the new working hours. Test the claim at 0.01 level of significance. The value of the test statistic is _____


-0.7814


0.7814


-20.3528


20.3528

Answer 1

To determine the test statistic for this hypothesis test, we can follow these steps:

1. **State the null and alternative hypotheses**:
- Null hypothesis (\(H_0\)): \(p \geq 0.33\) (33% or more of women are against the new working hours)
- Alternative hypothesis (\(H_1\)): \(p < 0.33\) (Less than 33% of women are against the new working hours)

2. **Determine the sample proportion (\( \hat{p} \))**:
- Number of women against the new hours = 45
- Sample size \(n\) = 150
- Sample proportion \( \hat{p} = \frac{45}{150} = 0.30 \)

3. **Determine the hypothesized population proportion (\(p_0\))**:
- \(p_0 = 0.33\)

4. **Calculate the test statistic using the formula for the z-test statistic**:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]

5. **Substituting values into the formula**:
\[
z = \frac{0.30 - 0.33}{\sqrt{\frac{0.33(1 - 0.33)}{150}}}
\]
First, we calculate \(p_0(1 - p_0)\):
\[
p_0(1 - p_0) = 0.33 \times 0.67 = 0.2211
\]
Now calculate the standard deviation:
\[
\sqrt{\frac{0.2211}{150}} \approx \sqrt{0.001474} \approx 0.0384
\]
Now we can find the z-value:
\[
z = \frac{0.30 - 0.33}{0.0384} = \frac{-0.03}{0.0384} \approx -0.78125
\]

6. **Final Result**:
The value of the test statistic is approximately \(-0.7814\).

Based on the choices provided, the answer is:

\(-0.7814\)