Answer:
a) The 95% of confidence intervals for the average spending
(23.872 , 32.128)
b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.
The null hypothesis is accepted
A survey of 25 young professionals statistically that the population mean is less than $30
Step-by-step explanation:
Step:-(i)
Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10
The sample size 'n' = 25
The mean of the sample x⁻ = $28
The standard deviation of the sample (S) = $10.
Level of significance ∝=0.05
The degrees of freedom γ =n-1 =25-1=24
tabulated value t₀.₀₅ = 2.064
Step 2:-
The 95% of confidence intervals for the average spending
([tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )[/tex]
[tex](28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )[/tex]
( 28 - 4.128 , 28 + 4.128)
(23.872 , 32.128)
a) The 95% of confidence intervals for the average spending
(23.872 , 32.128)
b)
Null hypothesis: H₀:μ<30
Alternative Hypothesis: H₁: μ>30
level of significance ∝ = 0.05
The test statistic
[tex]t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{28-30 }{\frac{10}{\sqrt{25} } }[/tex]
t = |-1|
The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.
The null hypothesis is accepted
Conclusion:-
The null hypothesis is accepted
A survey of 25 young professionals statistically that the population mean is less than $30
