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guajiro [1.7K]
3 years ago
11

Please help!!! Here’s a picture of my question

Mathematics
1 answer:
horrorfan [7]3 years ago
8 0

Answer:

option no. D

Step-by-step explanation:

i+34j+14k

hope it helps

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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 <em>X</em> = 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 - <em>α</em>)% 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 <em>z</em>-confidence interval.

The critical value of <em>z</em> 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 <em>z</em>-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 <em>z</em>-interval or <em>t</em>-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a <em>t</em>-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 <em>n</em> = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - <em>α</em>)% <em>t</em>-confidence interval is:

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

The critical value of <em>t</em> is:

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

*Use a <em>t</em>-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
3 years ago
Write the decimal 0.85 as a percentage
andreyandreev [35.5K]

Answer:

0.85 = 85%

Multiply 0.85 by 100 to get 85 percent

5 0
3 years ago
Read 2 more answers
A toy store ordered 26 crates of stuffed bears.
9966 [12]

Answer: C500

Step-by-step explanation: you would multiply 26 and 21 = 546 the round that out and you get C

4 0
4 years ago
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84 is 75% of what number?
Georgia [21]
75%=3/4
84/x=3/4
then you cross multiply
84*4=3x
26*4=x
104=x
<span>84 is 75% of 104</span>
5 0
3 years ago
Read 2 more answers
What is the slope of -1,-7 and 3,9
makkiz [27]

Answer:

<em>The slope is 4.</em>

Step-by-step explanation:

Use the slope formula.

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{-7 - 9}{-1 - 3}

m = \dfrac{-16}{-4}

m = 4

6 0
4 years ago
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