Suppose 216 subjects are treated with a drug that is used to treat pain and 40 of them developed nausea. Can you conclude that less than 25% of users develop nausea? Use the α = 0.02 level.
State the null and alternative hypotheses.
H0:
H1:

Find the critical value and sketch the rejection region.
Sketch of Rejection Region: One tailed – left

Critical value(s):

(If there are two critical values, list both values separated by a comma)

z_c=-2.326
Find the test statistic.

n=216
x=40
d=215
∝=0.02
p ̂=x/n=40/216=0.185
p_0=0.25
z=(p ̂-p_0)/√((p_0 (1-p_0))/n)=(0.185-0.25)/√((0.25(1-0.25))/216)=-2.206
Find the p-value.

p-value=0.014
Decide:       H0
P-value = 0.014 < 0.020 = significance level. We reject H0.

Conclusion: At α = 0.02 significance level, there      enough evidence to conclude that less than 25% of users develop nausea.

In 2012, the General Social Survey sampled 155 employed people and asked them how satisfied they were with their jobs. Of the 155 people sampled, 83 said that they were satisfied with their jobs. Can you conclude that more than 50% of employed people in the United States are satisfied with their jobs? Use the α = 0.14 level.
State the null and alternative hypotheses.
H0:
H1:

Find the critical value and sketch the rejection region.
Sketch of Rejection Region: One tailed – Right

Critical value(s):

(If there are two critical values, list both values separated by a comma)

z_c=1.476

Find the test statistic.

n=155
x=83
d=154
∝=0.14
p ̂=x/n=83/155=0.535
p_0=0.50
z=(p ̂-p_0)/√((p_0 (1-p_0))/n)=(0.535-0.50)/√((0.50(1-0.50))/155)=0.871

Find the p-value.

p-value=0.193

Decide:       H0
P-value = 0.193 > 0.140 = significance level. We do not reject H0.

Conclusion: At α = 0.14 significance level, there      enough evidence to conclude that more than 50% of employed people in the United States are satisfied with their jobs.

A popular blog reports that 55% of college students log in to Facebook on a daily basis. The Dean of Students at a certain university thinks that the proportion may be different at her university. She polls a simple random sample of 221 students, and 133 of them report that they log in to Facebook daily. Can you conclude that the proportion of students who log in to Facebook daily differs from 55%? Use the α = 0.08 level.
State the null and alternative hypotheses.
H0:
H1:

Find the critical value and sketch the rejection region.
Sketch of Rejection Region: Two-tailed

Critical value(s):

(If there are two critical values, list both values separated by a comma)

z_c=±1.759

Find the test statistic.

n=221
x=133
d=220
∝=0.08
p ̂=x/n=133/221=0.602
p_0=0.55
z=(p ̂-p_0)/√((p_0 (1-p_0))/n)=(0.602-0.55)/√((0.55(1-0.55))/221)=1.554

Find the p-value.

p-value=0.122

Decide:       H0
P-value = 0.122 > 0.080 = significance level. We do not reject H0.

Conclusion: At α = 0.08 significance level, there      enough evidence to conclude that the proportion of students who log in to Facebook daily differs from 55%.

A popular blog reports that 55% of college students log in to Instagram on a daily basis. The Dean of Students at a certain university thinks that the proportion may be different at her university. She polls a simple random sample of 278 students, and 170 of them report that they log in to Instagram daily. Can you conclude that the proportion of students who log in to Instagram daily differs from 55%? Use the α = 0.02 level.
State the null and alternative hypotheses.
H0:
H1:

Find the critical value and sketch the rejection region.
Sketch of Rejection Region: Two-tailed

Critical value(s):

(If there are two critical values, list both values separated by a comma)

z_c=±2.34

Find the test statistic.

n=278
x=170
d=277
∝=0.02
p ̂=x/n=170/278=0.612
p_0=0.55
z=(p ̂-p_0)/√((p_0 (1-p_0))/n)=(0.612-0.55)/√((0.55(1-0.55))/278)=2.078

Find the p-value.

p-value=0.039

Decide:       H0
P-value = 0.039 > 0.020 = significance level. We do not reject H0.

Conclusion: At α = 0.02 significance level, there      enough evidence to conclude that the proportion of students who log in to Instagram daily differs from 55%.

Type of Error: A        error could occur in this problem. The probability of this error occurring is          . We can reduce the probability of this error occurring by        .

Answer the following True or False:

If a researcher wants to test the claim that more than 25% of all students purchased their textbook online, then the null and alternative hypotheses are H0: p=0.25 and Ha: p>0.25.
false
true