Question

Out of a sample of 200 students, 75 indicated that they preferred chocolate ice cream to vanilla ice cream. |

a. |
Estimate the value
of the population proportion of those who preferred chocolate ice
cream. (Round the final answer to 3 decimal
places.) |

Estimated population proportion |

b. |
Compute
the standard error of the proportion. (Round the final answer to 4 decimal
places.) |

Standard error of the proportion |

c. |
Determine a 90%
confidence interval for the population proportion of those who
preferred chocolate ice cream. (Round the final answers to 3 decimal
places.) |

Confidence interval for the population proportion is between and . |

d. |
Interpret your findings. |

If 200 such intervals were determined, the population (Click to select)meanproportion would be included in about intervals. |

Just Need C and D as my answer for the practice lab is wrong. I want to know ehere im making the mistake.

Answer #1

a)

= 75 / 200 = **0.375**

b)

Standard error = Sqrt( ( 1 - ) / n)

= Sqrt( 0.375 * 0.625 / 200)

= **0.0342**

c)

90% confidence interval for p is

- Z * Sqrt( ( 1 - )/ n) < p < + Z * Sqrt( ( 1 - )/ n)

0.375 - 1.645 * 0.0342 < p < 0.375 + 1.645 * 0.0342

0.319 < p < 0.431

90% CI is (0.319 , 0.431)

Confidence interval for population proportion is between
**0.319** and **0.431**

d)

If 200 such intervals were determined, the population proportion would be included in

0.319 and 0.431.

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