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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

A study of teenage drivers found that 63% text while driving and 30% have a car accident in their first year of driving. 21% of

Kipish [7]

Answer:

33.33%

Step-by-step explanation:

63% of teenage drivers text while driving

21% of teenage drivers text while driving and have car accident in their first year

This means that a third of teenage drivers that text while driving get involved in a car accident during their first year, 33.33% is the probability then

6 0

3 years ago

During summer vacation, Erin read, on average, 4 pages per night. Once she returned to school, she averaged 3 pages per night. W

alexandr1967 [171]

Answer:

25% decrease.

Step-by-step explanation:

3/4=75%

100%-75%=25% decrease.

6 0

2 years ago

If a thermometer indicates 40 degrees Celsius, what is the temperature in degrees Fahrenheit?

Stells [14]

So we wan't to convert the temperature from Celsius scale to fahrenheit scale. The method fot that is this: First, 0 degrees Celsius or the starting point of Celsius scale is 32 degrees Fahrenheit. And the magnitude of a degree Fahrenheit is 1.8 times greater than the magnitude of a degree Celsius, or: 1F=1.8F. So now we can construct a function that will give us temperature in degrees Fahrenheit: T(F)=1.8T(C) + 32. We have T(C)= 40, and after we put that in our function we get: T(F)= 1.8*40 + 32 = 72 + 32 = 104 F. So 40 degrees Celsius is 104 degrees Fahrenheit.

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3 years ago

Andrew's mother reasoned that Chris probably took out a 30-year mortgage. Thirty years is a popular mortgage length because paym

melisa1 [442]

They would have increased because the 20% is bigger than the 5% and 5%=monthly housing expenses raising slow. 20%=monthly housing expenses raising higher.

6 0

2 years ago

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