Help please !! Solve -1.6r+5= -7.8

Find the critical value of x2 based on the given information H1 >3.5 n=14 a=0.05

egoroff_w [7]

Answer:

The degrees of freedom are given by:

$df = n-1=14-1=13$

The significance level is $\alpha=0.05$ and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

$\chi^2_{\alpha}= 22.362$

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

Step-by-step explanation:

We have the followign system of hypothesis:

Null hypothesis: $\sigma^2 \leq 3.5$

Alternative hypothesis: $\sigma^2 >3.5$

The degrees of freedom are given by:

$df = n-1=14-1=13$

The significance level is $\alpha=0.05$ and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

$\chi^2_{\alpha}= 22.362$

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

4 0

3 years ago

A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

So, the 95% confidence interval is (46.4, 53.6)

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

So, the 95% confidence interval is (47.9, 52.1)

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

Notice that the margin of error decreases when the sample size increases.

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