What is produced in photosynthesis

Total number of students surveyed = 200
Number of male students = 80
Number of female students = 200 - 80 = 120

Number of brown eyed male students = 60
Probability of a brown eyed male student = 60 / 80 = 0.75.

Since, <span>eye color and gender are independent, this means that eye color is not affected by the gender. Thus, we expect a similar probability of brown eye for female as we had for male.

Let the number expected of brown eyed females be x, then x / 120 = 0.75.

Thus, x = 120(0.75) = 90.

Therefore, the number female students surveyed expected to be brown eyed is 90.</span>

7 0

3 years ago

The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machi

Sonja [21]

Answer:

$t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349$

$p_v =P(t_{(72)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=88.8$ represent the sample mean

$s=26.6$ represent the sample standard deviation for the sample

$n=73$ sample size

$\mu_o =93$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ 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 true mean i lower than 93 min, the system of hypothesis would be:

Null hypothesis: $\mu \geq 93$

Alternative hypothesis: $\mu < 93$

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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (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:

$t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=73-1=72$

Since is a one side test the p value would be:

$p_v =P(t_{(72)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.

5 0

3 years ago