Answer: (16.65, 18.35)
Step-by-step explanation:
By considering the given information , we have
[tex]\overline{x}= 17.5[/tex] mm
s= 4.2 mm
n= 68
Significance level [tex]=\alpha=1-0.90=0.10[/tex]
Since population standard deviation is not given ,
Formula :
Confidence interval for population mean is given :-
[tex]\overline{x}\ \pm\ t_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
, where n = sample size.
[tex]t_{\alpha/2}[/tex] = Two-tailed t-value for significance level of [tex](\alpha)[/tex] and degree of freedom df= n-1.
[tex]s[/tex] = sample standard deviation.
Two-tailed t-value for significance level of [tex](0.10)[/tex] and degree of freedom df= 67:
[tex]t_{\alpha/2\ ,df}=t_{0.05,\ 67}=1.6679[/tex]
95% Confidence interval for population mean:
[tex]17.5\ \pm\ (1.6679)\dfrac{4.2}{\sqrt{68}}[/tex]
[tex]=17.5\ \pm\ (1.6679)\dfrac{4.2}{8.2462}[/tex]
[tex]=17.5\ \pm\ (1.6679)(0.5093255)[/tex]
[tex]=17.5\ \pm\ 0.8495[/tex]
[tex]=(17.5- 0.8495,\ 17.5+ 0.8495 )=(16.6505,\ 18.3495)\approx(16.65,\ 18.35)[/tex]
Hence, the 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel. = (16.65, 18.35)
