Question

An​ airline's public relations department says that the airline rarely loses​ passengers' luggage. It further claims that on those occasions when luggage is​ lost,

93 %93%

is recovered and delivered to its owner within 24 hours. A consumer group who surveyed a large number of air travelers found that only

116116

of

164164

people who lost luggage on that airline were reunited with the missing items by the next day. Does this cast doubt on the​ airline's claim? Explain.

Are the assumptions and the conditions to perform a​ one-proportion z-test​ met?

Yes

No

State the null and alternative hypotheses. Choose the correct answer below.

A.

H0​:

pequals=0.930.93

HA​:

pgreater than>0.930.93

B.

H0​:

pequals=0.930.93

HA​:

p not equals 0.93p≠0.93

C.

H0​:

pequals=0.930.93

HA​:

pless than<0.930.93

D.

The assumptions and conditions are not​ met, so the test cannot proceed.

Determine the​ z-test statistic. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

zequals=nothing

​(Round to two decimal places as​ needed.)

B.

The assumptions and conditions are not​ met, so the test cannot proceed.

Find the​ P-value. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

​P-valueequals=nothing

​(Round to four decimal places as​ needed.)

B.

The assumptions and conditions are not​ met, so the test cannot proceed.

Do the results of the survey cast doubt on the​ airline's claim? Explain. Choose the correct answer below.

A.

YesYes​,

since the null hypothesis is

not nbspnot rejected.

B.

YesYes​,

since the null hypothesis is

not nbspnot rejected.

C.

NoNo​,

since the null hypothesis is

nothingrejected.

D.

YesYes​,

since the null hypothesis is

nothingrejected.

E.

The assumptions and conditions are not​ met, so the test cannot proceed.

Answer 1

np = 164 * 0.93 = 152.52

n(1 - p) = 164 * (1 - 0.93) = 11.48

Since np > 10 and n(1 - p) > 10, so we can use one-proportion z-test.

Yes, the assumptions and the conditions to perform a one-proportion z-test met.

Option - B) H₀: P = 0.93

                  H_A: P 0.93

= 116/164 = 0.7073

The test statistic is

  

P-value = 2 * P(Z < -11.18)

            = 2 * 0 = 0

At 0.05 significance level, since the P-value is less than the significance level, so we should reject the null hypothesis.

Option - D) Yes, since the null hypothesis is rejected.