Pls help me with this :)

The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, al

Valentin [98]

Answer:

a) $p_v =P(t_{(9)}$

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \geq 0$

Alternative hypothesis: $\mu_y -\mu_x$

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679$

The next step is calculate the degrees of freedom given by:

$df=n-1=10-1=9$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(9)}$

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that $t_{\alpha/2}=3.25$

Now we have everything in order to replace into formula (1):

$-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469$

$-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069$

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

8 0

3 years ago

Jocelyn is training for a race by running several miles each day. she tracks her progress by recording her average speed in minu

Luden [163]

Finding the regression equation, her average speed on the 9th day should be expected to be of 6.92 minutes per mile.

<h3>How to find the equation of linear regression using a calculator?</h3>

To find the equation, we need to insert the points (x,y) in the calculator.

Researching the problem on the internet, the values of x and y are given as follows:

Values of x: 1, 2, 3, 4, 5, 6.

Values of y: 8.2, 8.1, 7.5, 7.8, 7.4, 7.5.

Hence, using a calculator, the equation for the average minutes per mile after t days is given by:

V(t) = -0.15143t + 8.28

Hence, for the 9th day, t = 9, hence the estimate is:

V(9) = -0.15143(9) + 8.28 = 6.92 minutes per mile.

More can be learned about regression equations at brainly.com/question/25987747

#SPJ1

4 0

1 year ago

For a fair coin, suppose you toss the coin 100 times.

Natasha_Volkova [10]

Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

For the binomial distribution, the parameters are given as follows:

n = 100, p = 0.5.

Hence the mean and the standard deviation of the approximation are given as follows:

$\mu = np = 100(0.5) = 50$ .

$\sigma = \sqrt{np(1-p)} = \sqrt{100(0.5)(0.5)} = 5$

Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is <u>one subtracted by the p-value of Z when X = 66.5</u>.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{66.5 - 50}{5}$

Z = 3.3

Z = 3.3 has a p-value of 0.9995.

1 - 0.9995 = 0.0005.

0.0005 = 0.05%

More can be learned about the normal distribution at brainly.com/question/4079902

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6 0

1 year ago

Connor is stuck in a traffic jam on a single-lane road. The traffic report tells him the traffic jam stretches for 5.5 miles. If

dedylja [7]

Answer:

2323

Step-by-step explanation:

According to the questions above, the report from the traffic jam says the traffics jam stretches for 5.5 miles

As Connor assumes that each car is about 12.5 long

The next step is to convert 5.5 miles into feet.

Since 1 mile = 5280

To convert 5.5 miles, 5.5 x 5280 = 29040 feet

To find the numbers of cars struck in traffic, we will have to divide 29040 by 12.5

29040/12.5 = 2323.3 ≅ 2323

5 0

3 years ago

Find the median and mean of the data set below:<br> 13, 6, 47,0

Readme [11.4K]

Answer:

Mean=16.5

Median=9.5

Step-by-step explanation:

4 0

2 years ago

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