In order to estimate the proportion of families who prefer female baby-sitters to male baby-sitters with an error of at most 8 percentage points and with at least 95 percent confidence, how large a sample should be taken?

## Testing for equality of meanFinal scores averaged 66 for the treatment group; the SD was 21. For the control group, the figures were 59 and 20. What do you conclude? |

## Determining the strength of given data using correlationDetermine the strength of the causal relationship between monthly sales and new home construction by using correlation. |

## Computing value of exponentially smoothed seriesUsing smoothing constant of w = .80, compute the value of exponentially smoothed series in 1983. |

## Prepare a frequency distributionPrepare a frequency distribution from data given below that describes how long it took a person to drive to work in work week. |

## Calculating the probability using z-scoreHow many ounces should they advertise on the box if they want to be sure that no more than 0.5% of the boxes are filled less than that advertised amount? |

## Confidence interval for the proportion-type of discomfortConstruct a 95% confidence interval for the proportion of all those who experience some type of discomfort. |

## Service charge for direct-deposit customersThe following data represent the service charge for direct-deposit customers if a customer's account falls below the minimum required $1,500 balance for a sample of 26 banks. |

## Mean of a normal distribution probabilityThe mean of a normal distribution probability is 400 pounds. The standard deviation is 10. What is the area between 415 pounds and the mean of 400 pounds? (Round to 4 decimals) |

## Computing mean of single-server queuing systemEntities arrive at single-server queuing system at average of 4 per hour and server can service average of 5 per hour. Mean in the queue is? |

## Finding confidence interval for average number of televisionIf possible, determine a 95% confidence interval for average number of television sets in all 50,000 households. If this is not possible, describe why not. |

## Length of time patron spend at the gymnasiumThe manager of a local gymnasium has determined that the length of time patron spend at the gymnasium is normal distributed with mean 80minutes and standard deviation 20 minutes. What is the patrons who spend more than 120 minutes at the gymnasium.. |

## Determining bivariate normal distributionCores on two tests have a bivariate normal distribution; and that correlation of 2 scores is 0.8. Determine the probability that sum of the student's scores on 2 tests will be greater than 200? |

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