Question

In: Statistics and Probability

The drying time of a type of paint under specified test conditions is known to be...

The drying time of a type of paint under specified test conditions is known to be normally distributed with mean value 75 min and standard deviation 9 min. Chemists have proposed a new additive designed to decrease average drying time. It is believed that drying times with this additive will remain normally distributed with σ = 9 min. Because of the expense associated with the additive, evidence should strongly suggest an improvement in average drying time before such a conclusion is adopted. Construct a suitable hypothesis test to check the feasibility of using the paint with the additive based on a sample of 25 painted specimens which resulted in an average drying time of 72.3 min.

 Use a Fixed Significance level-test and state your conclusions.

 Repeat the test with a P-Value approach test and state your conculsions.

 Construct an upper bound confidence interval and test the hypothesis based on it.

Solutions

Expert Solution

SOLUTION- WE SHALL USE MINITAB 16 FOR THE CALCULATIONS:

LET DENOTE THE AVERAGE DRYING TIME FOR THE OLD ADDITIVE AND DENOTE THE AVERAGE DRYING TIME FOR THE NEW ADDITIVE.

AS THE NEW ADDITIVE WAS DESIGNED TO REDUCE THE AVERAGE DRYING TIME, THE HYPOTHESIS WHICH COULD BE FRAMED AND TESTED HERE IS-

WE PERFORM A TWO SAMPLE-T TEST IN ORDER TO EXAMINE THE ABOVE STATED HYPOTHESIS-

STEPS- STAT> BASIC STATISTICS> TWO SAMPLE-T> ENTER THE SUMMARIZED DATA(SAMPLE SIZE,MEAN AND S.D)> UNDER 'OPTIONS', SET THE CONFIDENCE LEVEL 95.0( WE ASSUME 5% LEVEL OF SIGNIFICANCE), AND ALTERNATE AS 'LESS THAN'> CLICK OK.

1.) WE OBTAIN THE VALUE OF TEST STATISTIC AS, T= 1.06

NOW FOR A TWO SAMPLE-T TEST,UNDER EQUAL VARIANCE ASSUMPTION, D.F= 25+25-2 = 48

CRITICAL VALUE=

AS T-VALUE< CRITICAL VALUE, WE REJECT OUR NULL HYPOTHESIS AND CONCLUDE THAT THE NEW ADDITIVE REDUCES THE AVERAGE DRYING TIME

2.) THE OBSERVED P-VALUE IN THIS CASE IS 0.853

AS P-VALUE>LEVEL OF SIGNIFICANCE(0.05), WE ACCEPT THE NULL HYPOTHESIS AND CONCLUDE THAT THE AVERAGE DRYING TIME FOR THE NEW ADDITIVE IS NEARLY THE SAME AS THE PREVIOUS ONE AND THERE IS NO SIGNIFICANT DECREASE IN THE DRYING TIME.

3.) THE UPPER 95% CONFIDENCE INTERVAL IS 6.97

ALSO, THE ESTIMATE OF THE DIFFERENCE OBTAINED IS 2.70

WE KNOW THAT A 95% CONFIDENCE INTERVAL FOR A LEFT TAILED TEST IS  

AS THE CONFIDENCE INTERVAL CONTAINS THE ESTIMATE OF NULL HYPOTHESIS, THE RESULTS OF THE TESTING ARE NOT STATISTICALLY SIGNIFICANT.

**REMARK**- THE NECESSARY EXPLAINATIONS HAVE BEEN GIVEN ABOVE. IN CASE OF DOUBT, COMMENT BELOW. AND PLEASE LIKE, IF YOU LIKED THE SOLUTION.


Related Solutions

A) The drying time of a type of paint under specified test condition is known to...
A) The drying time of a type of paint under specified test condition is known to be normally distributed with an assumed mean value of 75 and a standard deviation of 9. Test the hypothesis H0: µ=75 vs. Ha: µ<75 from a sample of 25 observations. For a test procedure with α= 0.002 what is the probability of making a type 2 error when the true mean is 70, ß(70) (to 4 decimals)? B) The drying time of a type...
Consider a paint-drying situation in which drying time for a test specimen is normally distributed with...
Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 8. The hypotheses H0: μ = 75 and Ha: μ < 75 are to be tested using a random sample of n = 25 observations. (a) How many standard deviations (of X) below the null value is x = 72.3? (Round your answer to two decimal places.) 1.69 Correct: Your answer is correct. standard deviations (b) If x = 72.3, what is...
Consider a paint-drying situation in which drying time for a test specimen is normally distributed with...
Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 6. The hypotheses H0: μ = 75 and Ha: μ < 75 are to be tested using a random sample of n = 25 observations. (a) How many standard deviations (of X) below the null value is x = 72.3? (Round your answer to two decimal places.) Incorrect: Your answer is incorrect. standard deviations (b) If x = 72.3, what is the...
Consider a paint-drying situation in which drying time for a test specimen is normally distributed with...
Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 6. The hypotheses H0: μ = 74 and Ha: μ < 74 are to be tested using a random sample of n = 25 observations. (a) How many standard deviations (of X) below the null value is x = 72.3? (Round your answer to two decimal places.) standard deviations (b) If x = 72.3, what is the conclusion using α = 0.004?...
A paint manufacturer wants to determine the average drying time of a new interior wall paint....
A paint manufacturer wants to determine the average drying time of a new interior wall paint. If for sample 12 test areas of equal size he obtained a mean drying time of 66.3 minutes and a standard deviation of 8.4 minutes, construct a 95% confidence interval for the true mean μ is (___ , ___). PLEASE SHOW ALL YOUR WORK
A paint manufacturer wants to determine the average drying time of a new interior wall paint....
A paint manufacturer wants to determine the average drying time of a new interior wall paint. If for sample 12 test areas of equal size he obtained a mean drying time of 66.3 minutes and a standard deviation of 8.4 minutes, construct a 95% confidence interval for the true mean μ is (_____________ , _____________). (Show Work)
The mean drying time of a certain paint in a certain application is 12 minutes. A...
The mean drying time of a certain paint in a certain application is 12 minutes. A new additive will be tested to see if it reduces the drying time. One hundred specimens will be painted, and the sample mean drying time will be computed. Let μ be the mean drying time for the new paint. Assume the population standard deviation of drying times is 2 minutes. Assume that, in fact, the true mean drying time of the new paint is...
3. A paint manufacturing company claims that the mean drying time for its paints is not...
3. A paint manufacturing company claims that the mean drying time for its paints is not longer than 60 minutes. A random sample of 20 gallons of paints selected from the production line of this company showed that the mean drying time for this sample is 63.50 minutes with a sample standard deviation of 4 min utes. Assume that the drying times for these paints have a normal distribution. Test the claim that the mean drying time for its paints...
The drying time, in minutes, of two different brands of latex paint were recorded. In a...
The drying time, in minutes, of two different brands of latex paint were recorded. In a random sample of 6 units of brand 1, the mean average drying time is found to be 70.83 minutes with standard deviation of 15.09. However, for a random sample of 9 units of brand 2, the mean average drying time is found to be 79.33 minutes with standard deviation of 8.08. Assume normal populations. Note: Round all the final numbers to 2 decimal places....
A paint manufacturer wishes to compare the drying times of two different types of paint. Independent...
A paint manufacturer wishes to compare the drying times of two different types of paint. Independent random samples of 11 cans of Type A and 9 cans of Type B were selected and applied to a similar surface. The drying times were recorded. Type A had a mean drying time of 68.9 with a standard deviation of 0.5. Type B had a mean drying time of 70.6 with a standard deviation of 3. Do the data provide sufficient evidence at...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT