Question

Directions: Use the acronym PANIC to find the confidence intervals.

1.An SRS of 60 women showed that the average weight of a purse is 5 pounds with a standard deviation of 1.2 pounds. Find the 90% Confidence Interval for the actual average weight of purses.

2.A poll of 1256 households showed that 470 of them own a dog. Using this information find a 95% confidence interval.

Answer 1

a).given data are:-

Parameter : the population mean weight of purse ()

assumptions: simple random sample, normality assumption.

Name of interval : 1 sample t interval.

[ as here, sample sd is known we will t distribution ]

interval calculation :-

df = (n-1) = (60-1) =59

t critical value for df=59, alpha=0.10, both tailed test be:-

the 90% confidence interval be:-

Conclusion:-

we can be 90% confident that the true mean weight of purse will be within 4.74 pounds and 5.26 pounds.

b).given data are:-

Parameter : the population proportion of households who own a dog(p)

assumptions:

a). independence: the responses are independent of each other.

b).the sample is a random sample .

c).

so, the normality condition is satisfied.

Name of interval : 1 proportion z - interval.

interval calculation :-

z critical value for alpha=0.05, both tailed test be:-

the 95% confidence interval be:-

Conclusion:-

we can be 95% confident that the true proportion of households, who own a dog will be within 0.347 and 0.401

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