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-   -   Confidence interval for a standard deviation (https://www.askmehelpdesk.com/showthread.php?t=458899)

  • Mar 19, 2010, 01:09 PM
    bearmiss1
    Confidence interval for a standard deviation
    Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation, σ. Assume that the population has a normal distribution. Weights of eggs: 95% confidence; n = 22, x-bar = 1.78 oz, s = 0.47 oz


    a) 0.36 oz < σ < 0.65 oz
    b) 0.38 oz < σ < 0.63 oz
    c) 0.37 oz < σ < 0.61 oz
    d) 0.36 oz < σ < 0.67 oz
  • Mar 19, 2010, 01:12 PM
    bearmiss1
    Hypothesis Testing.
    Hypothesis Testing.
    A history teacher wants to determine if her teaching methods of this year have increased student scores above the previous year's mean of 80. She has sampled 16 students and found the sample mean to be 84.5 and the sample standard deviation to be 11.2.

    The null hypothesis, Ho, is: mu = 80.
    The alternative hypothesis, Ha, is: mu > 80.

    Determine if there is sufficient evidence to support whether the teacher has succeeeded. You will want to carry your calculations to three decimal places.

    a) t = 2.60, critical value is 1.61, reject Ho since t is greater than critical value.
    b) t = 2.31, critical value is 1.61, reject Ho since t is greater than critical value.
    c) t = 1.61, critical value is 2.31, accept Ho since t is less than critical value.
    d) t = 1.61, critical value is 2.60, accept Ho since t is less than critcal value.
  • Mar 19, 2010, 01:13 PM
    bearmiss1
    Hypothesis Testing
    Hypothesis Testing

    Use a 0.05 significance level to test the claim that peanut candies have weights that vary more than plain candies. The established standard deviation for the weights of plain candies is 0.373. A sample of 41 peanut candies was taken and the sample standard deviation was calculated to be 0.31. Assume the population is normally distributed.

    Null hypothesis: sigma <= 0.373
    Alternative hypothesis: sigma > 0.373 (Claim)

    a) The null hypothesis IS rejected. There IS NOT sufficient evidence to conclude the weights vary more for peanut candies.

    b) The null hypothesis IS rejected. There IS sufficient evidence to conclude the weights vary more for peanut candies.

    c) The null hypothesis IS NOT rejected. There IS NOT sufficient evidence to conclude the weights vary more for peanut candies.

    d) The null hypothesis IS NOT rejected. There IS sufficient evidence to conclude the weights vary more for peanut candies.
  • Mar 19, 2010, 01:14 PM
    bearmiss1
    Hypothesis Testing
    Hypothesis Testing
    A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 750 hours. A random sample of 46 light bulbs has a mean life of 735 hours with a standard deviation of 65 hours. Although the sample mean appears to be less than the claim, is there enough evidence to reject the manufacturer's claim using a significance level of 0.09?

    Null hypothesis: mu >= 750 (Claim)
    Alternative hypothesis: mu < 750

    You will want to take your calculations to 4 decimal places.

    a) Fail to reject Ho. There IS sufficient evidence to reject the claim that the mean bulb life is at least 750 hours.

    b) Reject Ho. There IS NOT sufficient evidence to reject the claim that the mean bulb life is at least 750 hours.

    c) Fail to reject Ho. There IS NOT sufficient evidence to reject the claim that the mean bulb life is at least 750 hours.

    d) Reject Ho. There IS sufficient evidence to reject the claim that the mean bulb life is at least 750 hours.
  • Mar 19, 2010, 01:16 PM
    bearmiss1
    Hypothesis Testing
    Hypothesis Testing

    It is claimed that 43% of home buyers find their real estate agent through a friend. A sample of 1600 home buyers reveals that 720 home buyers did find their real estate agent through a friend. At a 0.02 significance level, can you reject the claim that 43% of home buyers find their real estate agent through a friend?

    Null hypothesis: p = 0.43
    Alternative hypothesis: p <> 0.43

    a) FAIL TO REJECT the null hypothesis. There IS NOT sufficient evidence to reject the claim that 43% of home buyers find their real estate agent through a friend.

    b) REJECT the null hypothesis. There IS NOT sufficient evidence to reject the claim that 43% of home buyers find their real estate agent through a friend.

    c) FAIL TO REJECT the null hypothesis. There IS sufficient evidence to reject the claim that 43% of home buyers find their real estate agent through a friend.

    d) REJECT the null hypothesis. There IS sufficient evidence to reject the claim that 43% of home buyers find their real estate agent through a friend.
  • Mar 19, 2010, 01:17 PM
    bearmiss1
    Hypothesis Test for Two Samples
    Hypothesis Test for Two Samples
    A safety engineer records the braking distance of two types of tires. Each randomly selected sample has 35 tires. The results of the tests are shown in the table below:


    Type A Type B
    Number in Sample 35 35
    Sample Mean 43 feet 47 feet
    Sample Standard Deviation 4.9 feet 4.6 feet

    With a significance level of 0.10, can the engineer support the claim that the mean breaking distance is different for the two types of tires? Assume the sample are randomly selected and that the samples are independent.

    Null hypothesis: The breaking distance for both types is the same.
    Alternative hypothesis: The breaking distance both types is different. (Claim)

    a) FAIL TO REJECT the null hypothesis. The calculated test statistic LIES WITHIN the region(s) of rejection.
    b) FAIL TO REJECT the null hypothesis. The calculated test statistic DOES NOT LIE WITHIN the region(s) of rejection.
    c) REJECT the null hypothesis. The calculated test statistic LIES WITHIN the region(s) of rejection.
    d) REJECT the null hypothesis. The calculated test statistic DOES NOT LIE WITHIN the region(s) of rejection.
  • Mar 19, 2010, 01:18 PM
    bearmiss1
    Test Hypothesis Test for Two Samples
    Test Hypothesis Test for Two Samples
    A personnel director in a particular state claims that the mean annual income is greater in Jefferson County than in Washington County. A sample of 17 Jefferson County residents has a mean annual income of $41,700 and a standard deviation of $8,100. A sample of 8 Washington County residents has a mean annual income of $38,300 and a standard deviation of $5,100. Assume the population variances are not equal. At a 0.05 significance level, determine the critical value and test statistic.

    Null Hypothesis: mu1 - mu2 <= 0
    Alternative Hypothesis: mu1 - mu2 > 0 (Claim)

    a) Critical value = 1.895, Test Statistic = 1.564
    b) Critical value = 2.365, Test Statistic = 1.564
    c) Critical value = 1.895, Test Statistic = 1.275
    d) Critical value = 2.998, Test Statistic = 1.275
  • Mar 20, 2010, 04:06 AM
    galactus

    Looks like you have some homework problems to complete. Better get started. Show us some of your thoughts and we'll give a hand. These are straightforward hypothesis testing problems.
  • Mar 20, 2010, 08:09 PM
    morgaine300

    Oh my. The idea is to post a homework problem, and then post your attempts at the problem, or ask a question about the problem, then LEARN how to do the problem. Then apply that learning to your next problem and see what you can do with it, instead of just posting all the rest of the problems as well.

    You aren't going to learn anything by dumping all your homework at once onto a forum.

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