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3 years ago

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From the beginning of the day the temperature rose 15° so that by 4 pm it was 87°. What was the temperature at the start of the

day?
A)72°
B)77°
C)82°
D)102°

2 answers:

Pavlova-9 [17]3 years ago

5 0

Answer:

the answer is A)72°

Step-by-step explanation:

zubka84 [21]3 years ago

4 0

Your answer is A, 72°.

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2 years ago

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And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

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The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Assuming that the deviation is known we can express the margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65$

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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3 years ago