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Maru [420]
3 years ago
8

Of the 200 students surveyed at the local high school, 126 say that they prefer earlier start times for school. What is the mean

of the sampling distribution of the proportion of the students who prefer later start times for school
Mathematics
1 answer:
ra1l [238]3 years ago
5 0

Answer:

The mean of sampling distribution of the proportion of the students who prefer later start times for school

                        μₓ = p = 0.63

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given sample size 'n' =200

given data Of the 200 students surveyed at the local high school, 126 say that they prefer earlier start times for school

sample proportion

                        p = \frac{x}{n} = \frac{126}{200} = 0.63

The mean of sampling distribution of the proportion of the students who prefer later start times for school

                        μₓ = p = 0.63

Standard deviation of the proportion

                     S.D = \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.63(1-0.63)}{200} } = 0.034

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HELP! ASAP!<br> What is the measure of ∠A?<br><br><br> What is the measure of ∠B?
olganol [36]
The opposite angles of a parallelogram are equal.
Therefore
m∠A = m∠C, so that
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A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs
Makovka662 [10]

Answer:

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be​ 95% confident that the mean​ cost, μ​, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be​ 95% confident that the mean​ cost, μ​,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Hope this Helps!!!

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