In a certain large city, the mean birth weight of babies is 7.1 pounds with a standard deviation of 1.2 pounds. You believe that in one neighborhood in the city the mean birth weight is different from 7.1 pounds. You sample 95 births and find the mean weight of the sample to be 7.4 pounds. Can the claim be supported to a level of significance of α = .05, test the hypothesis?

No, the claim that the mean birth weight is 7.1 pounds cannot be supported at 0.05 significance level.

Step-by-step explanation:

A z-test is used to test the hypothesis because the population standard deviation is known.

Null hypothesis: The mean birth weight is 7.1 pounds.

Alternate hypothesis: The mean birth weight is not equal to 7.1 pounds.

The test is a two-tailed test because the alternate hypothesis is expressed with the inequality not equal to.

Test statistic (z) = (sample mean - population mean) ÷ (population sd/√n)

sample mean = 7.4 pounds

population mean = 7.1 pounds

population sd = 1.2 pounds

n = 95

z = (7.4 - 7.1) ÷ (1.2/√95) = 0.3 ÷ 0.123 = 2.44

At 0.05 significance level, the critical values from the standard normal distribution table are -1.96 and 1.96.

Conclusion:

Reject the null hypothesis because the test statistic 2.44 falls outside the region bounded by the critical values -1.96 and 1.96.

The claim cannot be supported at a 0.05 significance level.

answer: if you weigh 60 pounds on earth, on the moon you will weigh about 10 pounds. if you weigh 120 pounds on earth, you will weigh about 20 pounds on the moon. since weight is caused by the mass of two objects that are near each other, the bigger the objects, the more the force of gravity.

step-by-step explanation: if you weigh 60 pounds on earth, on the moon you will weigh about 10 pounds. if you weigh 120 pounds on earth, you will weigh about 20 pounds on the moon. since weight is caused by the mass of two objects that are near each other, the bigger the objects, the more the force of gravity.