90% = mean ± 1.645 SEm
SEm = SD/√n
Insert the values and calculate.
If it is claimed that the contents of soaps boxes sold weigh on average at least 20 ounces.
The distribution of weight is known to be normal;
We assume \sigma = 0.5σ=0.5 is known
n=30
sample mean weight = 18.5 ounces.
Test at the 10% significance level (\alpha = .10)(α=.10) the null hypothesis that the population mean weight is at least 20 ounces.
3 answers
Forget the above. I goofed. Try this.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
If it
is claimed that the contents
of soaps
boxes sold
weigh on average at least 20 ounces.
The distribution
of weight is known to be normal;
We
assume \sigma = 0.5σ=0.5 is known
n=30
sample
mean
weight =
18.5 ounces.
Test
at
the 10% significance level (\alpha = .10)(α=.10) the null
hypothesis that the population mean weight is at least 20
ounces.
is claimed that the contents
of soaps
boxes sold
weigh on average at least 20 ounces.
The distribution
of weight is known to be normal;
We
assume \sigma = 0.5σ=0.5 is known
n=30
sample
mean
weight =
18.5 ounces.
Test
at
the 10% significance level (\alpha = .10)(α=.10) the null
hypothesis that the population mean weight is at least 20
ounces.