What is the importance of DNA copying in reproduction ?
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Answer:
a) Null and alternative hypothesis
b) Test statistic t=-1.565
P-value = 0.0612
NOTE: the sample size is n=65.
c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
d) Null and alternative hypothesis
Test statistic t=-1.565
Critical value tc=-1.669
t>tc --> Do not reject H0
Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=65.
The sample mean is M=30.15.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
<u>Critical value approach</u>
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At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.
As the test statistic is greater than the critical value, it falls in the acceptance region.
The null hypothesis failed to be rejected.
The greatest number of sandwiches that Olivia can make is 12.
3 slices of cheese will be on each sandwich
<u>Solution:</u>
Given that, Olivia has 48 slices of ham and 36 slices of cheese to make sandwiches.
We have to find what is the greatest number of sandwiches that Olivia can make? And how many slices of cheese will be on each sandwich?
Let the number of sandwiches be "n"
Then, number of slices of ham per sandwich
And, number of slices of cheese per sandwich
As slices can’t be fractions, "n" should be a factor of 48 and 36. So now let us find H.C.F of 48 and 36 as we want greatest number of sandwiches that Olivia can make.
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48
The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, and 36
The common factors are those which appear on both factors of 48 and 36 are 1, 2, 4, 6, and 12.
Out of these common factors, 12 is the greatest
So, H.C.F of 48 and 36 is 12, which means n = 12
Then, number of cheese slices per sandwich
Hence, 12 sandwiches can be made with 3 slices of cheese per sandwich.