Question

Preliminary data analyses indicate that you can reasonably apply the z-interval procedure. Find a 90% confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Assume a population standard deviation of 190.0. Note: The sum of the data is 8279.

Answer 1

a) 90% confidence interval is (440.426, 591.574)

b) The range of values that, with a certain degree of confidence, represent the true value of the population parameter is known as the confidence interval. Option a is the right choice.

As per the given information,

Sum of the data is 8,772

The sample size is n =17.

Sample mean formula is as follows:

`\bar x=(1)/(n)\sum _(i=1)^n X_i`

`\bar x =(8,772)/(17)`

`\bar x` = 516

The population standard deviation is σ = 190.

The 100(1-a)% confidence interval for the population mean when the population standard deviation is known is given by,

CI = Point estimate+Margin of error

CI =
`\bar x` ±
`Z_(\alpha)/(2) (\sigma)/(√(n) )`

The confidence level is 90%.

Thus, the level of significance is a 0.10

1 - (
`(\sigma)/(2)`) = 0.95.

The absolute value of z-critical is 1.645.

CI =
`\bar x` ±
`Z_(\alpha)/(2) (\sigma)/(√(n) )`

CI = 516 ± 1.645
`((190)/(√(17) ) )`

CI = 516 ± 75.8045

CI = (440.426, 591.574)

The mean number of tongue flicks every 20 minutes for all young common lizards falls within the interval, with a confidence level of about 90%.
Option a is the right choice.

Answer 2

The CI of the 'true mean= [455, 609]

Explanation:

Please see attachment.