Answer:
The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
Step-by-step explanation:
The null hypothesis is:
The alternate hypotesis is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.
A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position.
This means that
Assume the population standard deviation is 6.2 weeks.
This means that
Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
We have to find the pvalue of Z, looking at the z-table, when . It if is lower than 0.05, it provides evidence.
has a pvalue of 0.0113 < 0.05.
The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
T=number of tiles bought
cost at store 1 is 0.79t+24
cost at store 2 is 1.19t
when will be equal cost?
0.79t+24=1.19t
minus 0.79t from both sides
24=0.4t
divide both sides by 0.4
60=t
answer is 60 tiles
9/27
there are infinite ratios equal to 9/27
one if the ratio is 1/3
For 4x^2y over 2x^8y^4 try 2 over x^6y^3