Can someone help me on #8?

Ashlyn had some Barbie dolls. Joslyn has 78 barbies. Rounded to the nearest ten, together they have 140 Barbie dolls. How many d

azamat

ashlyn has 60 barbie dolls

6 0

3 years ago

Read 2 more answers

7 over 9 divided by 3

GarryVolchara [31]

Fraction is 7/27 or 0.259

4 0

3 years ago

What is the simplified form of √45 A) √5 B) √9•5 C) 3√5 D)5√3

Kaylis [27]

The simplified form is C .

4 0

2 years ago

What dose 5d-10= using two word expressions

Naddik [55]

Maybe 5 multiplying or adding the value of d subtracting 10 equals

8 0

3 years ago

According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating

Natali [406]

Answer:

a) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

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

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

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:

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

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:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565$

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

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):

$\text{P-value}=P(t$

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.

Critical value approach

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.

6 0

2 years ago