Question

(S 9.2) Recall that a confidence interval for the sample mean can be calculated using the interval

*x*¯?*t**n*?1?*s**n*????*?*?*x*¯+*t**n*?1?*s**n*???

Thus, the margin of error is
*t**n*?1?*s**n*???

We can recover the margin of error from an interval constructed on the calculator using algebra.

Suppose a random sample of size 16 was taken from a normally
distributed population, and the sample standard deviation was
calculated to be *s* = 6.3. We'll assume the sample mean is
10 for convenience.

a) Calculate the margin of error for a 90% confidence interval for the population mean.

Round your response to at least 3 decimal places.

b) Calculate the margin of error for a 95% confidence interval for the population mean.

Round your response to at least 3 decimal places.

NOTE both these values are over 2. Suppose we want a smaller
margin of error.

c) Approximately how large of a sample size is needed to construct
a 90% confidence interval with a margin of error less than 1.5
given an estimate for the standard deviation of 6.3?

d) Approximately How large of a sample size is needed to construct
a 95% confidence interval with margine of error less than 1.5 given
an estimate for the standard deviation of 6.3?

Answer #1

) std error =std deviation/sqrt(n)=6.3/sqrt(10)

a)for (n-1=9) degree of freedom and 90% CI ; critical t =1.833

hence margin of error =t*std error=3.652

b)

for (n-1=9) degree of freedom and 95% CI ; critical t =2.262

hence margin of error =t*std error=4.506 ( please try 4.507 if this comes wrong due to rounding error)

c)

for 90 % CI value of z= | 1.645 |

standard deviation = | 6.30 |

margin of error E = | 1.5 |

required sample size n=(z/E)^{2
} = |
48.0 |

d)

for 95 % CI value of z= | 1.960 |

standard deviation = | 6.30 |

margin of error E = | 1.5 |

required sample size n=(z/E)^{2 }
= |
68.0 |

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