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3 years ago

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Mr. Cota’s family bought 4 sandwiches and 3 salads. They spent $45.50. Write an equation that represents this situation. Let x b

e the cost of a sandwich and y be the cost of a salad.

1 answer:

madreJ [45]3 years ago

8 0

45.50=4x+3y is the correct answer

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A) Two populations: American men and American women, as it estimated from the broad of the survey.

B) The proportion of men that sometimes do not drive safely while talking or texting on cell phone is 32%, and the proportion of women that sometimes do not drive safely while talking or texting on cell phone is 25%.

C) There is enough evidence to support the claim that the proportion of men and women who used their cell phone in an emergency differ significantly (P-value=0.014).

Step-by-step explanation:

A. We have two populations: American men and American women. This are the populations from which we want to infere the difference of means.

B. The estimation of the proportion is already shown in the question:

The proportion of men that sometimes do not drive safely while talking or texting on cell phone is 32%, and the proportion of women that sometimes do not drive safely while talking or texting on cell phone is 25%.

These are unbiased estimators of the populations proportions.

C. This is a hypothesis test for the difference between proportions.

The claim is that the proportion of men and women who used their cell phone in an emergency differ significantly.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1 (men), of size n1=643 has a proportion of p1=0.71.

The sample 2 (women), of size n2=643 has a proportion of p2=0.77.

The difference between proportions is (p1-p2)=-0.06.

$p_d=p_1-p_2=0.71-0.77=-0.06$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{457+495.11}{643+643}=\dfrac{952}{1286}=0.7403$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.7403*0.2597}{643}+\dfrac{0.7403*0.2597}{643}}\\\\\\s_{p1-p2}=\sqrt{0.0003+0.0003}=\sqrt{0.0006}=0.0245$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.06-0}{0.0245}=\dfrac{-0.06}{0.0245}=-2.45$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(z$

As the P-value (0.014) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of men and women who used their cell phone in an emergency differ significantly.

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