Matthew completed 5/8 of his workday. What decimal would also describe the portion of the workday he has finished? How can you u
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Answer:
The calculated value Z = 2.3084 > 1.96 at 5% level of significance
Null hypothesis is rejected
Alternative hypothesis is accepted
The hotel does not expects that 24% of all guests will request smoking rooms
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given the hotel takes a random sample of 223 guests. Out of this sample, 39 guests requested smoking rooms.
Sample proportion
p = x/n = 39/223 = 0.174
Given Population proportion 'P' = 0.24
Given random sample size 'n' = 223
Level of significance ∝ =0.05
The critical value Z₀.₀₅ = 1.96
<u><em>Null Hypothesis</em></u> (H₀) : P=0.24
<u><em>Alternative Hypothesis</em></u> (H₁) : P≠ 0.24
<u><em>Step(ii)</em></u>:-
<em>Test statistic </em>
|Z| = | -2.3084|= 2.3084
The calculated value Z = 2.3084 > 1.96 at 5% level of significance
Null hypothesis is rejected
Alternative hypothesis is accepted
The hotel does not expects that 24% of all guests will request smoking rooms
Answer:
<em>Hello your question is incomplete attached below is the complete question</em>
answer : There is not convincing statistical evidence to suggest that the proportion of customers who order dessert at each restaurant is not the same based on family classification. ( E )
Step-by-step explanation:
To arrive at this conclusion we will determine the Null and alternate hypothesis
<em>H0 : Number that orders dessert is same based on family classification given</em>
<em>Ha : Number that orders dessert is not the same based on family classification given </em>
from the question the p-value of Chi-square test is 0.092 > 0.05 hence we will fail to reject the null hypothesis. therefore we can conclude that
There is not convincing statistical evidence to suggest that the proportion of customers who order dessert at each restaurant is not the same based on family classification
