What is the perimeter of a baseball diamond 90 ft on each sode

A study by Consumer Reports showed that 64% of supermarket shoppers believesupermarket brands to be as good as national brands.

Elina [12.6K]

Answer:

$z=\frac{0.614 -0.64}{\sqrt{\frac{0.64(1-0.64)}{145}}}=-0.652$

The p value would be given by:

$p_v =P(z$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is significantly less than 0.64 at 5% of significance

Step-by-step explanation:

Information given

n=145 represent the random sample taken

X=89 represent the number of people who believed the supermarket brand was as good as the national brand

$\hat p=\frac{89}{145}=0.614$ estimated proportion of people who believed the supermarket brand was as good as the national brand

$p_o=0.64$ is the value to test

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is less than 0.64, the system of hypothesis are.:

Null hypothesis: $p \geq 0.64$

Alternative hypothesis: $p < 0.64$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info we got:

$z=\frac{0.614 -0.64}{\sqrt{\frac{0.64(1-0.64)}{145}}}=-0.652$

The p value would be given by:

$p_v =P(z$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is significantly less than 0.64 at 5% of significance

5 0

3 years ago

Ryan spent $3.25 on lunch every day, Monday through Friday. If he had $20 at the start of the week, how much money did he have l

shusha [124]

Monday through Friday is 5 days.

Multiply the cost of lunch by number of days:

3.25 x 5 = $16.25

Subtract the total he spent on lunch from what he started with for money:

20 - 16.25 = 3.75

He had $3.75 left.

6 0

3 years ago