Answer:
90% Confidence interval = (18.423, 20.577)
95% Confidence interval = (18.211, 20.789)
Step-by-step explanation:
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.
Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be foumd with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Sample Mean = 19.5
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.
To find the critical value from the t-tables, we first find the degree of freedom and the significance level.
Degree of freedom = df = n - 1 = 65 - 1 = 64.
Significance level for 90% confidence interval
(100% - 90%)/2 = 5% = 0.05
significance level for 95% confidence interval
(100% - 95%)/2 = 2.5% = 0.025
t (0.05, 64) = 1.669
t (0.025, 64) = 1.998
Standard error of the mean = σₓ = (σ/√n)
σ = standard deviation of the sample = 5.2
n = sample size = 65
σₓ = (5.2/√65) = 0.645
90% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 19.5 ± (1.669 × 0.645)
CI = 19.5 ± 1.076505
90% CI = (18.423495, 20.576505)
90% Confidence interval = (18.423, 20.577)
95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 19.5 ± (1.998 × 0.645)
CI = 19.5 ± 1.28871
95% CI = (18.21129, 20.78871)
95% Confidence interval = (18.211, 20.789)
Hope this Helps!!!
