Company A: A 2-column table with 5 rows. The first column is labeled hours with entries 5, 12, 20, 29, 42. The second column is
2 answers:
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Answer:
(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716
(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814
Step-by-step explanation:
We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;
Mean, = 30.05 inches and Standard deviation,
= 0.2 inches
Let X = A sheet selected at random from the population
Here, the standard normal formula is ;
Z = ~ N(0,1)
(a) <em>The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65) </em>
P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)
P(X < 30.65) = P( <
) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425
= 0.9985
P(X <= 30.25) = P( <=
) = P(Z <= 1) = 0.84134
Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .
(b)<em> Let Y = Standard Normal Variable is given by N(0,1) </em>
<em> Which means mean of Y = 0 and standard deviation of Y = 1</em>
So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)
P(Y < 3.2) = P( <
) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688
= 0.99931
P(Y <= 0.3) = P( <=
) = P(Z <= 0.3) = 0.61791
Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .
Answer:
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a two sided test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.