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Bas_tet [7]
3 years ago
15

A study is performed in a large southern town to determine whether the average weekly grocery bill per four-person family in the

town is significantly different from the national average. The national average grocery bill is $100 (standard deviation unknown). A random sample of 25 families is taken in the town, and the mean is found to be 103.23 with a standard deviation of 8.82.
For each item below, please show your work and/or explain your reasoning to receive full credit.
a). State the null and alternative hypotheses for this situation.
b). Assuming a 0.05 level of significance, compute the appropriate test statistic.
c). Find the critical t value. Can you reject the null hypothesis? Explain/justify your response using numbers and words
Mathematics
2 answers:
OLga [1]3 years ago
4 0

Answer:

The average weekly grocery bill per four-person family in the town is significantly different from the national average.          

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $100

Sample mean, \bar{x} = $103.23

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = $8.82

First, we design the null and the alternate hypothesis

H_{0}: \mu = 100\text{ dollars}\\H_A: \mu \neq 100\text{ dollars}

We use Two-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{103.23- 100}{\frac{8.82}{\sqrt{25}} } = 1.831

Now,

t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = \pm 2.063

Since,                

t_{stat} < t_{critical}

We accept the null hypothesis.

We accept the null hypothesis and determine that the average weekly grocery bill per four-person family in the town is significantly same to the national average.

IrinaK [193]3 years ago
4 0

Answer:

From the given problem statement

Average=$100

mean=103.23

standard deviation=8.82

Significance level=0.05

confidence level=1-0.05=99.95

The null hypothesis is given below

H0=100

H0>103.23

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(Intercept) 415.113     82.517   5.031 0.000709 ***  

x1            -6.593      4.859 -1.357 0.207913    

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---  

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  

Residual standard error: 24.45 on 9 degrees of freedom  

Multiple R-squared: 0.768,     Adjusted R-squared: 0.7164  

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a).  y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

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e).  newdata=data.frame(x1=21.3, x2=43)

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1 81.03364 43.52379 118.5435

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1 81.03364 14.19586 147.8714

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