Golf Cart Industry
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1999 Taylor Dunn electric cart golf / industrial / yard - 4 passenger 36 volt | ![]() |
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US $1,750.00 | 5h 50m |
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SOLENOID RELAY GOLF CART & INDUSTRIAL APPLICATION 36V | ![]() |
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US $23.25 | 8d 1h 44m |
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NEW SOLENOID WINCH MOTOR GOLF CART MARINE INDUSTRIAL | ![]() |
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US $24.90 | 20d 17h 25m |
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NEW SOLENOID SWITCH Golf Cart Industrial Cont. Duty 24V | ![]() |
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US $21.95 | 21d 18h 31m |
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EZ GO Golf Cart part brake drum 5 lug industrial veh. | ![]() |
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US $59.99 | 21d 1h 24m |
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Club Car Golf Cart Carburetor Carryall, Industrial FE350 1992-1997 *New In Box* | ![]() |
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US $109.99 | 29d 21h |
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Golf Cart Industry

Which method is faster?
The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?
sdenison has the right idea, but a bit incorrect in the application of the details.
You are given n observations and the sample standard deviation and the sample mean. In order to apply the sample standard deviation for this analysis, you have to consider the estimate of the population standard deviation, which is = the square root of the sample variance divided by n.
So, population SD = sqrt( (s^2) / 24) = sqrt ( (2.7^2) / 24) = 0.551135.
Now, I am interpreting "10% level of significance" to mean that there is a 90% probability that an observation will be less than or equal to a certain value. The "certain value" is the sample mean + 1.28 * population standard deviation. The 1.28 comes from the standard normal distribution table (z-score tables), where 90% of the distribution lies below +1.28 SD's above the mean.
So, our 10% level of significance figure should be 40.6 + (1.28 * 0.551135) = 41.30545. Since our original time (42.3) is higher than this 90% figure (41.30545), we can say that at the 90% significance level, the original time is not the same as the new time.
(Statistics is a touchy subject - you can only reject a hypothesis or fail to reject a hypothesis. You can't "accept a hypothesis". In this case, we reject the hypothesis that the original time and the new times are the same.)
If the 41.30505 answer does not match what you have, then try dividing the sample variance by n-1 (which is 23) instead of n (which is 24). Dividing by n-1 gives you an unbiased estimator of the standard deviation, which your textbook might be trying to emphasize.
Inside Look - Augusta's Golf-Cart Industry Suffers in Recession - Bloomberg
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1999 Taylor Dunn electric cart golf / industrial / yard - 4 passenger 36 volt | ![]() |
![]() |
US $1,750.00 | 5h 50m |
| Powered by phpBay Pro |
![]() |
![]() |
SOLENOID RELAY GOLF CART & INDUSTRIAL APPLICATION 36V | ![]() |
![]() |
US $23.25 | 8d 1h 44m |
![]() |
NEW SOLENOID WINCH MOTOR GOLF CART MARINE INDUSTRIAL | ![]() |
![]() |
US $24.90 | 20d 17h 25m |
![]() |
NEW SOLENOID SWITCH Golf Cart Industrial Cont. Duty 24V | ![]() |
![]() |
US $21.95 | 21d 18h 31m |
![]() |
EZ GO Golf Cart part brake drum 5 lug industrial veh. | ![]() |
![]() |
US $59.99 | 21d 1h 24m |
![]() |
Club Car Golf Cart Carburetor Carryall, Industrial FE350 1992-1997 *New In Box* | ![]() |
![]() |
US $109.99 | 29d 21h |
| Powered by phpBay Pro |
![]() |
![]() |
Club Car Golf Cart Carburetor Carryall, Industrial FE350 1992-1997 *New In Box* | ![]() |
![]() |
US $109.99 | 29d 21h |
| Powered by phpBay Pro |
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1967 Stevens Utility Electric Car Industrial Truck Brochure Golf Cart | ![]() |
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US $11.99 | 1d 2h 47m |
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Suncast GO3216 Golf Organizer
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DescriptionShow your golf clubs the love they deserve by storing them properly. The Suncast golf organizer has room for two golf bags, and a protective foam strip ensures that the shafts don't get scratched. It also includes three shelves and a five-inch bin you can use for storing shoes, balls, tees, and so on... |
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Universal Power Group 45966 Sealed Lead Acid Battery
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Zen Golf: Mastering the Mental Game
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DescriptionZen Golf...Learn The Keys To Mastering The Mental Game! The best players know that great golf comes from confidence and concentration-the ability to stay in the present and block out distraction. Achieving a clear mind is also at the heart of Buddhist philosophy and practice... |
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Viair 00073 70P Heavy Duty Portable Compressor |
DescriptionCompact and Powerful portable air compressor. |
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E-Z Red S101 Battery Hydrometer
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DescriptionTemperature compensatingObtain direct, accurate readingsDesigned so that anyone can see the state of the battery acidWide range of 1.100 - 1.300 sg.Unbreakable, E-Z to read, economicalUsed by battery manufacturers worldwide Weight: 4 oz. |
Help with this statistics problem and pls show the work!?
The management of White Industries is considering a new method of assembling its golf
cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The
mean assembly time for a random sample of 24 carts, using the new method, was 40.6minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of
significance, can we conclude that the assembly time using the new method is faster?
Small Sample Hypothesis Test for mean:
In order for this test to be valid the data must come from a normal population. If this is not the case then this test is not valid and other methods, such as a randomization test or permutation test should be used.
Assuming the normality assumption is valid to test the null hypothesis
H0: μ ≤ Δ or
H0: μ ≥ Δ or
H0: μ = Δ
Find the test statistic t = (xbar - Δ ) / (sx / √ (n))
where xbar is the sample average
sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.
n is the sample size
and t follows the Student t distribution with n - 1 degrees of freedom. We use the Student t distribution to account for the uncertainty in the estimate of the variance.
As the degrees of freedom approach infinity the Student t converges in probability to the Standard Normal. In most cases the values of the percentiles of the Student t are close enough to the Standard Normal when the degrees of freedom are greater than 30. This is the source of the empirical rule of thumb that samples of size > 30 have a mean that is normally distributed. Keep that in mind as well, for these hypothesis tests we are assuming the mean is normally distributed. This assumption is easy to verify if the data is normally distributed. The Central Limit Theorem accounts of all other means.
The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.
H1: μ > Δ; p-value is the area to the right of t
H1: μ < Δ; p-value is the area to the left of t
H1: μ ≠ Δ; p-value is the area in the tails greater than |t|
If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true.
If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.
The hypothesis test in this question is:
H0: μ ≥ 42.3 vs. H1: μ < 42.3
The test statistic is:
t = ( 40.6 - 42.3 ) / ( 2.7 / √ ( 24 ))
t = -3.084543
The p-value = P( t_ 23 < t )
= P( t_ 23 < -3.084543 )
= 0.00261788
Since the p-value is less than the significance level of 0.1 we reject the null hypothesis and conclude the alternate hypothesis μ < 42.3 is true.






















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