Question

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. Construct a 90% confidence interval for the population mean.

Answer 1

Answer:

90% confidence interval for the population mean is $149.01,\ 150.99$

Step-by-step explanation:

Given that, in a random sample of 60 computers, the mean repair cost was $150, with the population standard deviation is $36.

Here,

$n=60,\mu=150,\sigma=36$

In case of 90% confidence interval, $Z=1.645$

We know that, the confidence interval will be,

$=\mu\pm Z\dfrac{\sigma}{n}$

Putting all the values,

$=150\pm 1.645\times \dfrac{36}{60}$

$=150\pm 0.987$

$=149.01,\ 150.99$

Answer 2

Given that:
x_bar=150, σ=36, n=60
The 90% confidence interval will be found as follows:
+/-1.65(36/sqrt(60))
=+/-1.65(4.64758)
=+/-7.6685
Thus our answer will be:
150+/-7.6685
=157.6685 or 142.3315