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Aneli [31]
3 years ago
7

A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)

Find and interpret the 95% confidence interval for the average spending. (b) Based on your answer in part (a), can you conclude statistically that the population mean is less than $30? Explain

Mathematics
1 answer:
skad [1K]3 years ago
8 0

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

<u>Step-by-step explanation</u>:

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

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 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

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

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

<u>Conclusion</u>:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

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