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Sati [7]
3 years ago
12

A simple random sample of 85 is drawn from a normally distributed population, and the mean is found to be 146, with a standard d

eviation of 34. Which of the following values is outside the 99% confidence interval for the population mean? Use the table below to help you answer the question.
Confidence Level
90%
95%
99%
z*-score
1.645
1.96
2.58

Remember, the margin of error, ME, can be determined using the formula M E = StartFraction z times s Over StartRoot n EndRoot EndFraction.
the value of 135 because it is not greater than 136.5
the value of 137 because it is greater than 136.5
the value of 138 because it is less than 153.2
the value of 154 because it is greater than 153.2
Mathematics
2 answers:
makvit [3.9K]3 years ago
7 0

Answer:

35

Step-by-step explanation:

This is the answer because it is not greater than 135

Grace [21]3 years ago
5 0

Answer:

The correct answer is a. the value of 35 because it is not greater than 136.5

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Here,

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A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 u
USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Gas mileage A: Mean 36, standard deviation 6, sample of 50:

So

\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

So

\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314

Distribution of the difference:

Mean:

\mu = \mu_A - \mu_B = 36 - 42 = -6

Standard error:

s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = zs = 1.96*1.4142 = 2.77

The margin of error is of 2.77 mpg

C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

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1 is the answer hope this helps
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