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In an examination of holiday spending (known to be normally distributed) of a sample of 16 holiday shoppers at a local mall, an average of $54 was spent per hour of shopping. Based on the current sample, the standard deviation is equal to $21. Find a 90% confidence interval for the population mean level of spending per hour

Respuesta :

Answer:

(45.6965,63.2035)

Step-by-step explanation:

Given that in an examination of holiday spending (known to be normally distributed) of a sample of 16 holiday shoppers at a local mall, an average of $54 was spent per hour of shopping.

i.e. [tex]\bar x = 54\\ n = 16\\s = 21\\std error = \frac{21}{\sqrt{16} } \\=5.25[/tex]

Margin of error = t critical * std error=9.2035

df = 15

(Here since sample std dev is known we use t critical value)

Confidence interval lower bound = [tex]54-9.2035 = 45.6965[/tex]

Upper bound = [tex]54+9.2035=63.2035[/tex]

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