Can someone help please?

A microwave manufacturing company has just switched to a new automated production system. Unfortunately, the new machinery has b

baherus [9]

Answer:

Null hypothesis: $\mu \leq 6$

Alternative hypothesis: $\mu > 6$

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$

$df=n-1=36-1=35$

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

$p_v =P(t_{(35)}>2)=0.0267$

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=6.5$ represent the mean time for the sample

$s=1.5$ represent the sample standard deviation for the sample

$n=36$ sample size

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05,0.1$ 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 mean is higher than 6 days, the system of hypothesis would be:

Null hypothesis: $\mu \leq 6$

Alternative hypothesis: $\mu > 6$

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:

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$

Critical value and P-value

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

$df=n-1=36-1=35$

In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates $\alpha$ on the right. Using the significance level of 0.05 we got:

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

And using the significance of 0.1 we got

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

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

$p_v =P(t_{(35)}>2)=0.0267$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.

6 0

3 years ago

Katelyn is training for a marathon. This week she ran 6 miles less than 2 times the number of miles she ran the previous week. S

Kobotan [32]

Answer:

The system of Equations =

r = 2p - 6........ Equation 1

r + p = 42 ........Equation 2

Hence:

the number of miles she ran this week = r = 26 miles

the number of miles she ran the previous week = p = 16 miles

Step-by-step explanation:

Let the number of miles:

the number of miles she ran this week = r

the number of miles she ran the previous week = p

Katelyn is training for a marathon. This week she ran 6 miles less than 2 times the number of miles she ran the previous week.

r = 2 × p - 6 miles

r = 2p - 6........ Equation 1

She ran a total of 42 miles in the two weeks.

r + p = 42 miles

r + p = 42 ........Equation 2

Which system of equations can be used to find the number of miles she ran this week, r, and the number of miles she ran the previous week,p?

The system of Equations =

r = 2p - 6........ Equation 1

r + p = 42 ........Equation 2

We substitute 2p - 6 for r in Equation 2

2p - 6 + p = 42

3p = 42 + 6

3p = 48

p = 48/3

p = 16 miles

Solving for r

r = 2p - 6........ Equation 1

r = 2(16) - 6

r = 32 - 6

r = 26 miles

Hence:

the number of miles she ran this week = r = 26 miles

the number of miles she ran the previous week = p = 16 miles

7 0

3 years ago

Is my answer correct??

Studentka2010 [4]

Answer:

The correct answer is the first choice: Hinge Theorem.

Explanation:

The Hinge Theorem compares the lengths of two sides of two triangles, given that the other two pairs of sides are congruent and the included angles are different.

The theorem states that if two sides of a triangle are congruent to two sides of another triangle and the included angle of one triangle is greater than the included angle of the other triangle, then length of the third side of the first triangle is greater than the length of the third side of the second triangle.

The figures shows triangles UVT and STV

The angle UVT is greater than the angle STV.

The sides ST and VU are congruent, as the small vertical marks indicate.

The side TV is a common side of both triangles.

Then, you already have that two sides of the triangle STV are congruent to two sides of triangle UVT and the angle UVT is greater than the angle STV.

Hence, by the HInge Theorem the side TU is greater than side SV.

3 0

3 years ago

Point K is rotated 90°. The coordinate of the pre-image point K was (2, –6) and its image K’ is at the coordinate (−6, −2). Find

aalyn [17]

Answer:

A clockwise rotation.

Step-by-step explanation:

The rule for 90 degree clockwise rotation is (x,y) turns into (y,-x) therefore if you use your numbers (2,-6) and then use the 90 degree clockwise rotation rule it becomes (-6,-2)

5 0

3 years ago

Eleven graduate students have applied for three available teaching assistantships. In how many ways can these assistantships be

kykrilka [37]

3men, 1 women. for the first 2. and 2 men, 1 women for the third

8 0

2 years ago

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