Question

An article states that movie attendance declined in 2005. A survey found that 720 of 1000 randomly selected adult Americans preferred to watch movies at home rather than at a movie theater. Is there convincing evidence that the majority of adult Americans prefer to watch movies at home? Test the relevant hypotheses using a 0.05 significance level. (For z give the answer to two decimal places. For P give the answer to four decimal places.)

z=

p=

Answer 1

The calculated test statistic z is approximately 4.40, and the corresponding p-value is extremely small. With a significance level of 0.05, we reject the null hypothesis, concluding that the majority prefer watching movies at home.

To test whether there is convincing evidence that the majority of adult Americans prefer to watch movies at home, we can use a hypothesis test for proportions.

Null Hypothesis
`(\(H_0\))`:

`\(p \leq 0.5\)` (the majority do not prefer to watch movies at home)

Alternative Hypothesis
`(\(H_1\))`:

p > 0.5 (the majority prefer to watch movies at home)

Test Statistic (z):

`\[ z = \frac{\hat{p} - p_0}{\sqrt{(p_0(1 - p_0))/(n)}} \]`

where

-
`\(\hat{p}`) is the sample proportion (720/1000),

-
`\(p_0\)` is the hypothesized proportion under the null hypothesis (0.5),

- n is the sample size (1000).

Calculation:

`\[ z = \frac{0.72 - 0.5}{\sqrt{(0.5(1 - 0.5))/(1000)}} \]`

`\[ z \approx \frac{0.22}{\sqrt{(0.25)/(1000)}} \]`

`\[ z \approx (0.22)/((0.5)/(10)) \]`

`\[ z \approx (0.22)/(0.05) \]`

`\[ z \approx 4.40 \]`

P-value:

Using a standard normal distribution table, the p-value for
`\(z \approx 4.40\)` is extremely small.

Conclusion:

Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. There is convincing evidence that the majority of adult Americans prefer to watch movies at home.