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3 years ago

Suppose 100 users of a social networking site are chosen for a sample, but

Mathematics

1 answer:

serious [3.7K]3 years ago

4 0

Answer:

i think its d

Step-by-step explanation:

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A certain railway company claims that its trains run late 5 minutes on the average. The actual times (minutes) that 10 randomly

iragen [17]

Answer:

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 5 minutes

Sample mean, $\bar{x}$ = 9.130 minutes

Sample size, n = 10

Alpha, α = 0.01

Sample standard deviation, s = 1.4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 5\text{ minutes}\\H_A: \mu \neq 5\text{ minutes}$

We have to construct 99% confidence interval.

99% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.01} = \pm 3.249$

$9.130 \pm 3.249(\dfrac{1.4}{\sqrt{10}} ) = 9.130 \pm 1.4383 = (7.6917,10.5683)$

99% Confidence interval: (7.6917,10.5683)

Since the population mean does not lie in he calculated confidence interval, thus, we fail to accept the null hypothesis and accept the alternate hypothesis.

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

6 0

3 years ago

92. How much material was required for

erma4kov [3.2K]

answer:92+92=184 184×5=920

5 0

3 years ago

Please help me!!!!!!!!!!!!!!!!!!!!

daser333 [38]

26.25 divided by 3 is 8.75 so the answer is B, just divide the about of money by the number of hours and you have your answer

6 0

2 years ago

Lipids provide much of the dietary energy in the bodies of infants and young children. There is a growing interest in the qualit

garri49 [273]

Answer:

This assumption is that the distribution of polyunsaturated fat % of each if these four regimes must be with equal variances as well as uniform.

The null hypothesis

H0: μ1 = μ2 = μ3 = μ4

The alternate hypothesis:

H1: at least 2 means are unequal

First we calculate the grand mean

= 1/54[15(42.9)+17(43.1)+8(43.7)+14(43.9)]

= (643.5 +732.7 + 349.6 + 614.6)/54

= 2340.4/54

= 43.341

Sum of squared treatment

= [15(42.9-43.341)²+17(43.1-43.341)²+8(43.7-43.341)²+14(43.9-43.341)²]

= 9.3104

Mean square of treatment

= SST/I-1

= 9.3104/4-1

= 9.3104/3

= 3.1035

Error sum of squared

= (15-1)*(1.3)² + (17-1)*(1.5)² + (8-1)*(1.2)² + (14-1)*(1.2)²

= 88.46

Error mean square

MSE = 88.46/54-4

= 1.7692

Test statistic

= 3.1035/1.7692

= 1.75

4 0

3 years ago

Plllssss help me its the last one

DerKrebs [107]

Well I think step 1 is wrong she is suppossed to add 32.75 +32.75 then add she would have 65.50 after she would have to add 70.99 her results form that would be 136.49 after she would need to subtract 12.25 then she ends up with 124.24 and 1/4 of that is 31.06 therefore what she put in her savings was 31.06 I hope this helps!

3 0

4 years ago