In a survey, 11 people were asked how much they spent on their child's last birthday gift. the results were roughly bell-shaped with a mean of $46 and standard deviation of $14. construct a confidence interval at a 90% confidence level.

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus .

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

,

where

For this confidence interval:

Hence the width of the 90% confidence interval is

.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

your answer is the frist one

hop i