Question

Eric wants to estimate the percentage of elementary school children who have a social media account. He surveys 450 elementary school children and finds that 280 have a social media account. Identify the values needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval.

z0.10 |
z0.05 |
z0.025 |
z0.01 |
z0.005 |

1.282 | 1.645 | 1.960 | 2.326 | 2.576 |

Use the table of common *z*-scores above.

- Round the final answer to three decimal places.

$p'=\ $ sigma sub p prime is equal to $\sigma_{p'}\ =\ $ z sub alpha over 2 is equal to $z_{\frac{\alpha}{2}}\ =\ $ (, )

Answer #1

Sample proportion = 280 / 450 = 0.622

99% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.622 - 2.576 * sqrt [ 0.622 * (1 - 0.622) / 450] < p < 0.622 + 2.576 * sqrt [ 0.622 * (1 - 0.622) / 450]

0.563 < p < 0.681

99% CI is **( 0.563 , 0.681 )**

Suppose the manager of a shoe store wants to determine the
current percentage of customers who are males. How many customers
should the manager survey in order to be 98% confident that the
estimated (sample) proportion is within 5 percentage points of the
true population proportion of customers who are males? z0.10 z0.05
z0.04 z0.025 z0.01 z0.005 1.282 1.645 1.751 1.960 2.326 2.576 Use
the table of values above?

Suppose the manager of a shoe store wants to determine the
current percentage of customers who are males. How many customers
should the manager survey in order to be 92% confident that the
estimated (sample) proportion is within 5 percentage points of the
true population proportion of customers who are males?
z0.10
z0.05
z0.04
z0.025
z0.01
z0.005
1.282
1.645
1.751
1.960
2.326
2.576
Use the table of values above.

Suppose the manager of a shoe store wants to determine the
current percentage of customers who are males. How many customers
should the manager survey in order to be 99% confident that the
estimated (sample) proportion is within 5 percentage points of the
true population proportion of customers who are males?
z0.10
z0.05
z0.04
z0.025
z0.01
z0.005
1.282
1.645
1.751
1.960
2.326
2.576
Use the table of values above.

Suppose the manager of a shoe store wants to determine the
current percentage of customers who are males. How many customers
should the manager survey in order to be 80% confident that the
estimated (sample) proportion is within 6 percentage points of the
true population proportion of customers who are males?
z0.10
z0.05
z0.04
z0.025
z0.01
z0.005
1.282
1.645
1.751
1.960
2.326
2.576
Use the table of values above.

Suppose the manager of a shoe store wants to determine the
current percentage of customers who are males. How many customers
should the manager survey in order to be 95% confident that the
estimated (sample) proportion is within 6 percentage points of the
true population proportion of customers who are males?
z0.10
z0.05
z0.04
z0.025
z0.01
z0.005
1.282
1.645
1.751
1.960
2.326
2.576
Use the table of values above.
Provide your answer below:

Suppose the manager of a shoe store wants to determine the
current percentage of customers who are males. How many customers
should the manager survey in order to be 95%
confident that the estimated (sample) proportion is within
5 percentage points of the true population
proportion of customers who are males?
Z0.10
Z0.05
Z0.04
Z0.025
Z0.01
Z0.05
1.282
1.645
1.751
1.960
2.326
2.576

The lengths of text messages are normally distributed with a
population standard deviation of 4 characters and an unknown
population mean. If a random sample of 27 text messages is taken
and results in a sample mean of 23 characters, find a 95%
confidence interval for the population mean. Round your answers to
two decimal places.
z0.10
z0.05
z0.04
z0.025
z0.01
z0.005
1.282
1.645
1.751
1.960
2.326
2.576

The lengths of text messages are normally distributed with a
population standard deviation of 3 characters and an unknown
population mean. If a random sample of 29 text messages is taken
and results in a sample mean of 30characters, find a 92% confidence
interval for the population mean. Round your answers to two decimal
places
z0.10
z0.05
z0.04
z0.025
z0.01
z0.005
1.282
1.645
1.751
1.960
2.326
2.576
select the correct answer below:
(28.56,31.44)
(29.29,30.71)
(29.08,30.92)
(28.70,31.30)
(29.02,30.98)
(28.91,31.09)

The lengths of text messages are normally distributed with a
population standard deviation of 3 characters and an unknown
population mean. If a random sample of 26 text messages is taken
and results in a sample mean of 29 characters, find a 98%
confidence interval for the population mean. Round your answers to
two decimal places. z0.10 z0.05 z0.04 z0.025 z0.01 z0.005 1.282
1.645 1.751 1.960 2.326 2.576 You may use a calculator or the
common z-values above. Select the...

1.)The yearly incomes, in thousands, for 24 random married
couples living in a city are given below. Assume the yearly incomes
are approximately normally distributed. Use Excel to find the 95%
confidence interval for the true mean, in thousands. Round your
answers to three decimal places and use increasing order. Yearly
Income 59.015 58.962 58.935 58.989 58.997 58.970 59.000 59.014
59.001 59.003 58.992 58.926 59.032 58.958 59.093 58.955 59.003
58.952 59.057 59.056 59.074 59.128 59.001 59.007
2.)In a recent questionnaire...

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