Steve rounds his favorite number to the nearest hundred to get 400.

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

3 years ago

Use rounding numbers to estimate the sum of 171 + 727

Umnica [9.8K]

171 rounds to 200, and 727 rounds to 700. 700+200= 900.

The answer is 900.

5 0

3 years ago

Read 2 more answers

If an egg is thrown up and off the roof of a 100 foot building with an initial upward velocity of 10 feet per second, how long u

Deffense [45]

The equation of the height as a function of time for this case is given by:
h (t) = - 16t ^ 2 + 10t + 100
By the time the egg hits the ground we have:
-16t ^ 2 + 10t + 100 = 0
We look for the roots of the polynomial:
t1 = -2.2069555463432966
t2 = 2.8319555463432966
As it is about time, we use the positive root:
t = 2.83 s
Answer:
it hits the ground at:
t = 2.83 s

8 0

4 years ago

Lnes made9 necklaces with 60 beads on each necklaces. How many beads did she use in all

Gala2k [10]

Made 9 necklaces.
60 beads per necklace.
How many beads for all the 9 necklaces?

60 × 9 = 540

9 necklaces = 540 beads

3 0

3 years ago

Kim’s hair grows a half inch per month. How many months will it take for her hair to grow 12 inches?

Contact [7]

Answer:

24 months

Step-by-step explanation:

Cross-multiply

0.5 inches -------------- 1 month

12 inches ---------------- x months

0.5/12 = 1/x ===> 0.5x = 12 ===> x = 12/0.5 ===> x = 24 months

3 0

3 years ago