Question

Georgianna claims that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. What is the 90% confidence interval for the average number of years students take piano lessons in this city?

Answer 1

Answer: $(3.75,\ 5.45)$

Step-by-step explanation:

Given : Sample size of  children : n= 20

Degree of freedom = df =n-1 =20-1=19

Sample mean years of piano lessons : $\overline{x}=4.6$

Sample standard deviation : $s= 2.2$

Confidence level : $1-\alpha= 0.90$

Significance level : $\alpha= 1-0.90=0.10$

Since population standard deviation is unavailable, then

Confidence interval for the population mean :

$\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$  (1)

Using t-distribution table , we have

Critical value = $t_{\alpha/2, df}=t_{0.05 , 19}=\pm1.7291$

The 90% confidence interval for the average number of years students take piano lessons in this city will be :

$4.6\pm (1.7291)\dfrac{2.2}{\sqrt{20}}$  (Substitute the values in (1))

$4.6\pm (1.7291)(0.491935)$

$\approx4.6\pm 0.85$

$(4.6- 0.85,\ 4.6+ 0.85)$

$(3.75,\ 5.45)$

Hence, the 90% confidence interval for the average number of years students take piano lessons in this city =$(3.75,\ 5.45)$