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4 years ago

6

Ann's watch runs 2 minutes per hour too slow. Beth's watch runs 1 minute per hour too fast. For example, if both watches were co

rrect at 12p.m

1 answer:

marysya [2.9K]4 years ago

5 0

Please add the rest of the excercise in order to help you! Have a nice a day! :)

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3 years ago

Read 2 more answers

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

8 0

4 years ago

The results of the first 100 students who voted are represented in the table. There are still 50 more students left to vote. Bas

Volgvan

Answer:

1-50

Step-by-step explanation:

5 0

3 years ago

Based on historical data in Oxnard college, we believe that 37% of freshmen do not visit their counselors regularly. For this ye

slega [8]

Answer:

A sample size of 1031 is required.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

37% of freshmen do not visit their counselors regularly.

This means that $\pi = 0.37$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

You would like to be 98% confident that your estimate is within 3.5% of the true population proportion. How large of a sample size is required?

A sample size of n is required.

n is found when M = 0.035. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.035 = 2.327\sqrt{\frac{0.37*0.63}{n}}$

$0.035\sqrt{n} = 2.327\sqrt{0.37*0.63}$

$\sqrt{n} = \frac{2.327\sqrt{0.37*0.63}}{0.035}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.37*0.63}}{0.035})^2$

$n = 1030.4$

Rounding up:

A sample size of 1031 is required.

4 0

3 years ago