• The average depth of the Hudson Bay is 305 feet. Climatologists were interested in seeing if the effects of warming and ice melt were affecting the water level. Fifty- five measurements over a period of weeks yielded a sample mean of 306.2 feet. The population variance is known to be 3.57. Can it be concluded at the .05 level of significance that the average depth has increased? (Use the Traditional method

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 305

For the alternative hypothesis,

H1: µ > 305

This is a right tailed test

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 305

x = 306.2

σ = 3.57

n = 55

z = (306.2 - 305)/(3.57/√55) = 2.49

Test statistic = 2.49

The calculated test statistic is 2.49 for the right tail and - 2.49 for the left tail

Since α = 0.05, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.05/2 = 0.025

The z score for an area to the left of 0.025 is - 1.96

For the right, α/2 = 1 - 0.025 = 0.975

The z score for an area to the right of 0.975 is 1.96

In order to reject the null hypothesis, the test statistic must be smaller than - 1.96 or greater than 1.96

Since - 2.49 < - 1.96 and 2.49 > 1.96, we would reject the null hypothesis.

Therefore, at 5% level of significance, there is sufficient evidence to conclude that the average depth has increased.

answer: c i think

step-by-step explanation:

the forst and 3rd

step-by-step explanation: