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Bess [88]
3 years ago
10

The color reference provided by the automotive manufacture is 44.21%. Does this known value match the measured value at the 95%

confidence level
Mathematics
1 answer:
Elis [28]3 years ago
4 0

Answer:

This known value does not match the measured value at the 95% confidence level.

The 95% confidence interval for value is calculated as 42.92 to 43.78%,

Step-by-step explanation:

The complete question is:

Police have a hit-and-run case and need to identify the brand of red auto paint. The percentage of iron oxide, which gives paint its red color, was determined to be 43.35 ± 0.33% by one method of analysis using five measurements. The color reference provide by the automotive manufacture is 44.21%. What can you conclude about whether the sample matches the reference at a 95% confidence level?

Student's t test is used to compare the two values .

Here n=5

Degrees of freedom =ν= n-1= 5-1=4

The value from t- table for 4 d.f at 95 % confidence interval is 2.776

The test statistic is

x`± t ∝/2(v). S/√n

43.35 ± (2.776)(0.33)/√5

= 43.35 ± 0.43

= 43.78, 42.92

Here the 95% confidence interval for μ calculated is 42.92 to 43.78%, where as color reference provided by the automotive manufacture is 44.21%.

This known value does not match the measured value at the 95% confidence level.

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Step-by-step explanation:

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A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

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Associative
☆is the order.
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