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2 years ago

The following data are the daily number of minutes of smartphone use for a random sample of 8 students at your institution:

Mathematics

1 answer:

Nadya [2.5K]2 years ago

8 0

Answer:

We fail to reject the Null ;

(88.323, 130.677)

Step-by-step explanation:

Given the data :

117, 156, 89, 72, 116, 125, 101, 100

H0 : μ ≥ 100

H1: μ < 100

From the data: using calculator :

Mean, x = 109.5

Standard deviation, s = 25.33

The test statistic:

(x - μ) ÷ s/Sqrt(n)

(109.5 - 100) ÷ 25.33/sqrt(8)

Test statistic = 1.06

Since, same size is small, we use t test ;

Using the Pvalue calculator :

T at df = 8 - 1 = 7, α = 0.05

Pvalue = 0.161

Pvalue > α

0.161 > 0.05 ; Hence, we fail to reject H0.

B.)

C. I = mean ± Tcritical *s/√n

Tcritical = 2.365 at 95% and df = 7

109.5 ± 2.365(25.3264/√8)

109.5 ± 2.365(8.9542)

(109.5-21.1768 ; 109.5+21.1768)

(88.323, 130.677)

In conclusion true mean is in the confidence interval, so it is consistent.

You might be interested in

Select all quantities that are proportional to the diagonal length of a square.

horsena [70]

Answer:

B. Perimeter of a square and

C. Side length of a square

Step-by-step explanation:

if n= side length of square then

Area of square is $n^{2}$
Perimeter of a square is 4×n
diagonal length of a square is $\sqrt{2}$ × n

Thus,

Perimeter of square can be expressed as

$2\sqrt{2}$ ×diagonal length of a square

Side length of a square can be expressed as

$\frac{1}{\sqrt{2} }$ ×diagonal length of a square

but Area of square is

$\frac{1}{\sqrt{2} }$ ×n×diagonal length of a square

As a Result, Area of square is also dependent of the value n, wheras in other cases it is a proportion of diagonal length of a square

5 0

3 years ago

A laptop producing company also produces laptop batteries, and claims that the batteries

gregori [183]

Answer:

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours. (P-value = 0).

The null and alternative hypothesis are:

$H_0: \mu=4\\\\H_a:\mu< 4$

Step-by-step explanation:

The question is incomplete: To test this claim a sample or population standard deviation is needed.

We will estimate that the sample standard deviation is 2 hours, and use a t-test to test that claim.

NOTE (after solving): The difference between the sample mean and the mean of the null hypothesis is big enough to reject the null hypothesis, even when we have a sample standard deviation of 3.5 hours, which can be considered bigger than the maximum standard deviation for the sample.

This is a hypothesis test for the population mean.

The claim is that the batteries power the laptops for significantly less than 4 hours.

Then, the null and alternative hypothesis are:

$H_0: \mu=4\\\\H_a:\mu< 4$

The significance level is 0.05.

The sample has a size n=500.

The sample mean is M=3.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{500}}=0.0894$