Question

In 2010, it was reported that teens sent on average 3,300 texts per month. It is believed that the number of texts sent has increased. In fact, a recent random sample of 60 teens showed that they send an average 3,500 texts per month(X¯=3,500 textsX¯=3,500 texts)  with a sample standard deviation of 500 texts (s=500 textss=500 texts). Assume that the random variable, number of texts sent by teens (denoted by X), is normally distributed.

Is this sufficient statistical evidence to show that teens are now sending on average more than 3,300 texts per month?

Question 16

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Specify the null and alternative hypotheses.

Select one:

a. H(0): μ≥3,300μ≥3,300   versus H(a): μ<3,300μ<3,300

b. H(0): μ≤3,300μ≤3,300 versus H(a): μ>3,300μ>3,300

Question 17

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The standard error (SE) of X¯X¯ is

Select one:

a. 64.55

b. 0.24

c. 0.76

d. 3.01

Question 18

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The test statistics value is

Select one:

a. 2.08

b. 0.53

c. 0.66

d. 3.10

Question 19

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The p-value is

Select one:

a. 0.999

b. 0.973

c. 0.001

d. 0.301

Question 20

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At α=0.01 and using the p-value

Select one:

a. We do not reject H(0)

b. We reject H(0) in favor of H(a)

Answer 1

Answer 16

For testing the hypothesis we need to perform a z-test for one mean, with known population standard deviation.

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ ≤ 3300 and Ha: μ > 3300 (Option B)

Answer 17

Standard Error (SE) = σ/SQRT(n)

Here, σ = 500 and n = 60

SE = 500/SQRT(60)

SE = 64.55 (Option A)

Answer 18

Test Statistics

The z-statistic is computed as follows:

The test statistics value is 3.10 (Option D)

Answer 19

The p-value corresponding to z = 3.10 for right tailed test is 0.001 (Option C)

(P value was obtained using online calculator. Screenshot given below)

Answer 20

Since p-value (0.001) < α (0.01), we reject null hypothesis (Option B)