Pls Hurry Need Help Asap I Will Give Brainliest

10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the

Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

$H_0: \mu = 0$

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

$H_1: \mu > 0$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

In this problem, the parameters are given as follows:

$\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50$ .

Hence, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}$

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by $t^{\ast} = 2.4$ .

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

5 0

1 year ago

You wish to find a 95% confidence interval for the mean number of times men change channels with a remote control during a comme

gogolik [260]

For margin of error to be a maximum of 3:
3 = z*(SD / sqrt(n)), where z is the z-score, SD is the standard deviation and n is the sample size.
z = 1.96 for a 95% confidence interval, and we are given SD = 15.
3 = 1.96*15/sqrt(n)
sqrt(n) = 9.8
n = 96.04 ~ 96 commercials.

4 0

3 years ago

A towns population grew from 7500 people to 8210 over the span of 10 years. What is the rate of growth in people per year for th

lisov135 [29]

Answer: 71 people per year

Step-by-step explanation:

8210 - 7500 = 710

710 ÷ 10 = 71.

Rate of growth in people per year for those 10 years = 71 people per year.

7 0

2 years ago

The distribution of the average amount of sleep per night gotten by college students is roughly bell-shaped with mean 412 minute

kirza4 [7]

Answer:

$P(X$

$P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(412,68)$