Purchase Solution

Standard Deviation, Significance Level

Not what you're looking for?

Ask Custom Question

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?

Purchase this Solution
Solution Summary

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?

Solution Preview

we can conduct a t-test for the case:
<br>Ho: M=40.6 or the mean assembly time is 40.6 minutes
<br>Ha: M>40.6 or the mean assembly time is ...

Purchase this Solution
Free BrainMass Quizzes
Measures of Central Tendency

Tests knowledge of the three main measures of central tendency, including some simple calculation questions.

Measures of Central Tendency

This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.

Know Your Statistical Concepts

Each question is a choice-summary multiple choice question that presents you with a statistical concept and then 4 numbered statements. You must decide which (if any) of the numbered statements is/are true as they relate to the statistical concept.

Terms and Definitions for Statistics

This quiz covers basic terms and definitions of statistics.

View More Free Quizzes
Related BrainMass Solutions

  • Test of hypothesis for proportions, standard deviation

    Use a .01 significance level to test the claim that for men without college degrees in that town, incomes have a higher standard deviation when a random sample of 22 men resulted in incomes with a standard deviation of 925$ 1.

  • Sample Size, Margin of Error, Confidence Level

    the population standard deviation from sample standard deviation we will use t distribution t at the 0.1 level of significance and 9 degrees of freedom (=n-1) and2 tailed test= 1.8331 E = margin of error= t*sx= 2.0868

  • P-value for the hypothesis test

    Data Hypothesized mean= 220.0 Standard deviation= 17.0 Sample mean= 230.8 Sample size= 12 Significance level= 0.05 0r 5% Population standard deviation (Known, Not Known)= Known

  • Is there evidence that the mean cost for student textbooks at the University of China is above $310?

    Null Hypothesis: Mean= $310 Alternative Hypothesis: Mean> $310 Mean=M =$310.00 Standard deviation =s=$43.20 sample size=n=20 sx=standard error of mean=s/square root of n=9.6598= ( 43.2 /square root of 20) Significance level=a (alpha)

  • Test of Hypothesis -Testing Gas Mileage Claims

    Note: Since significance level for testing the hypothesis is not mentioned we can assume it to be 0.05 Data Hypothesized mean= 28 mpg Standard deviation= 4.2 mpg Sample mean= 26.7 mpg Sample size=

  • Test of hypothesis

    A previously determined population standard deviation is $200. Test to determine if the mean of all account balances is significantly different from $1,000. Use a 0.05 level of significance.

  • Test of hypothesis

    Data Hypthesized mean= 3 minutes Standard deviation= 1 minute Sample mean= 2.75 minute Sample size= 50 Significance level= 0.05 0r 5% Population standard deviation (Known, Not Known)= Known

  • The solution to Test of hypothesis

    Data Hypthesized mean= 400 Standard deviation= 6 Sample mean= 407 Sample size= 12 Significance level= 0.01 0r 1% Population standard deviation (Known, Not Known)= Not Known

  • Confidence interval for mean

    level= 99% Therefore Significance level=alpha (a) = 1% =100% -99% No of tails= 2 2 tails because we are calculating the confidence interval Since sample size= 18 < 30 and we are using sample standard deviation to estimate

  • Hypothesis test: sample of 100

    Details Z Test of Hypothesis for the Mean Data Null Hypothesis m= 80 Level of Significance 0.01 Population Standard Deviation 12 Sample Size 100 Sample Mean 78.5 Intermediate Calculations Standard Error of the

View More Related Solutions