Calculating Standard Deviations
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5.The credit manager of a discount sports equipment store believes that the average age of his charge account customers is 30 years. His deputy contends that the average age is less than 30 years. A random sample of 100 charge account customers showed the mean age as 27 years with a standard deviation of 10 years. Do these data support the deputy's belief? What is the p-value?
6.Exhibit 8-18 According to a newspaper report, the average monthly rent for one-bedroom apartments in Quant City is $790. A local Realtor suspects that the report overstated the average monthly rent. She took a random sample of 50 one-bedroom apartments and recorded an average monthly rent of $825 with a standard deviation of $120.Refer to Exhibit 8-18. What is the p-value for testing if the report overstated the average monthly rent? (Since the sample size is sufficiently large, use z.)
9.How large a sample would be needed to estimate, within 24 hours, the mean burning time of incandescent lamps with no greater risk than one chance in 20 of being wrong. The standard deviation of burning time is 200 hours.
a. 250
b. 357
c. 189
d. 267
e. 301
10. The mean monthly expenditure on gasoline per household in Middletown is determined by selecting a random sample of 36 households. The sample mean is $68 with a random sample standard deviation of $17. Calculate a 90% confidence interval for the mean monthly expenditure on gasoline per household in Middletown. INTERPRET.
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Solution Summary
The expert calculates the standard deviations. The p-value for testing is computed.
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5.The credit manager of a discount sports equipment store believes that the average age of his charge account customers is 30 years. His deputy contends that the average age is less than 30 years. A random sample of 100 charge account customers showed the mean age as 27 years with a standard deviation of 10 years. Do these data support the deputy's belief? What is the p-value?
Ho: µ>=30
Ha: µ<30
This is a left tailed z test.
Test value z=(27-30)/(10/sqrt(100))=-3
P(z<-3)= 0.00135
Therefore, we reject null hypothesis. These data supports the deputy's belief and the P value is 0.00135. ...
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