Respuesta :
Answer:
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: yes, the claim can be supported.
Step-by-step explanation: This is a question under hypothesis testing of one sample mean.
The null hypothesis is given below as
H0: u = 7.1
Which implies that the mean birth weight of babies is 7.1
The alternative hypothesis is given below as
H1 : u > 7.1 ( sample mean (x) = 7.4 )
From the alternative hypothesis relative to the null, we can see that the test is a one tailed test and upper tailed.
Hence, we have our parameters as listed below as
Population mean (u) = 7.1
Sample mean (x) = 7.4
Population standard deviation (σ) = 1.2
Sample size = 95
Level of significance = 5%.
We will be using a z test to get our test statistics, this is because sample size is greater than 30 ( n= 95) and population standard deviation is given.
The z score is given below as
Z = x - u/ (σ)/√n
By substituting our parameters, we have that
Z = 7.4 - 7.1 / (1.2/√95)
Z = 0.3/ 0.1231.
Z = 2.44
We need to get the probabilistic value (p- value) attached to this z score, and that can only be done by using a standard normal distribution table.
Because our test is upper tailed, we will be looking for the value of z> 2.44 on the table.
From the standard normal distribution table, z > 2.44 = 0.0073.
We need to compare this with the level of significance to know whether to accept or reject the initial claim.
If p > 0.05, we would reject the null hypothesis because a large value of p means we have a higher chance of committing a type one error ( note level of significance is the probability of commuting a type one error)
If p < 0.05, we would accept the null hypothesis because we have a less chance of commuting a type one error.
From our solution, p = 0.0073, which is lesser than 0.05, hence we accept the initial claim that the average mean weight of babies is 7.1
