Question

In an examination of purchasing patterns of shoppers, a sample of 36 shoppers revealed that they spent, on average, $50 per hour of shopping. Based on previous years, the population standard deviation is $4.80 per hour of shopping. Assuming that the amount spent per hour of shopping is normally distributed, calculate a 90% confidence interval.

Answer 1

Answer:

the 90% confidence interval is ( 48.684  , 51.316  )

Step-by-step explanation:

Given that :

the sample size = 36

Sample Mean = 50

standard deviation = 4.80

The objective is to calculate a 90% confidence interval.

At 90% confidence interval ;

the level of significance = 1 - 0.9 = 0.1

The critical value for $z_{\alpha/2} = z_{0.1/2}$

$= z_{0.05}$ = 1.645

The standard error S.E = $\dfrac{\sigma}{\sqrt{n}}$

=$\dfrac{4.8}{\sqrt{36}}$

$=\dfrac{4.8}{6}$

= 0.8

The Confidence interval level can be computed as:

$\bar x \ \pm z \times \ \dfrac{ \sigma }{\sqrt{n}}$

For the lower limit :

$\bar x \ - z \times \ \dfrac{ \sigma }{\sqrt{n}}$

$=50 \ - 1.645 \times \ \dfrac{ 4.8 }{\sqrt{36}}$

$=50 \ - 1.645 \times \ 0.8 }}$

=50 - 1.316

= 48.684

For the upper limit :

$\bar x \ - z \times \ \dfrac{ \sigma }{\sqrt{n}}$

$=50 \ + 1.645 \times \ \dfrac{ 4.8 }{\sqrt{36}}$

$=50 \ + 1.645 \times \ 0.8 }}$

=50 + 1.316

= 51.316

Thus, the 90% confidence interval is ( 48.684  , 51.316  )