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The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The

Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(c) The necessary assumption is that the population is normally distributed.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let X = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

(a)

The general form of a (1 - α)% confidence interval is:

$CI=SS\pm CV\times SD$

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a z-confidence interval.

The critical value of z for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the confidence interval as follows:

$CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)$

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

$37\pm 0.1=SS\pm CV\times SD$

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

$P(-1$

The percentage of z-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the z-interval or t-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a t-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For n = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - α)% t-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The critical value of t is:

$t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262$

*Use a t-table for the critical value.

The 95% confidence interval is:

$CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)$

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

3 0

2 years ago

Eight figures are below. Which ones are convex polygons?

Nataly_w [17]

Answer:

A, C, D, F, G. I believe so?

7 0

3 years ago

A garden is 14 1/2 ft by 13 1/4 ft. What is the area

Marianna [84]

The area means to multiply the length by the width.

let's solve.

14 1/2 x 13 1/4.
convert our fractions into decimals.

14 1/2 = 14.50
13 1/4 = 13.25

Now we can multiply properly.

14.50 x 13.25 = 192.125
since our shape is 2-Dimensional, our units are ^2.

Your area is:
192.25 ft^2

3 0

3 years ago

The sum of three consecutive numbers is 72. What is the middle number?

Mkey [24]

Answer:

48

Step-by-step explanation:

Assuming that the numbers went in a pattern instead of a random order it would be 48 because 48 divided by 3 would be 24. 24 would be the first number though, so it would be 48 because 24 plus 24 is 48.

Hope this helps, have a great day!

6 0

3 years ago

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Find the volume of this square based pyramid. 10 in 12 in [? ]in3

zaharov [31]

Answer:

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