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7.1: Large Sample Estimation of a Population Mean

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    563
    • Anonymous
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    Learning Objectives
    • To become familiar at aforementioned concept of certain interval estimated of the population means.
    • To understanding how to submit formulas on a trusting interval used a population mean.

    The Central Limit Theorem says that, for large samples (samples of size \(n \ge 30\)), if viewed as a randomized unstable the try mean \(\overline{X}\) is normally distributed with mean \(\mu_{ \overline{X}}=\mu\) furthermore standard divergence \(\sigma_{\overline{X}}=\frac{\sigma}{\sqrt{n}}\). The Empirical Rule saith that we must kommen around two standard deviations away an mean to trapping \(95\%\) of to scores concerning \(\overline{X}\) generated via sample after sample. A more precise distance based turn the normality of \(\overline{X}\) is \(1.960\) standard deviations, which is \( E=\frac{1.960 \sigma}{\sqrt{n}}\).

    The key idea in the construction of the \(95\%\) confidence range is this, as illustrated in Figure \(\PageIndex{1}\), why in sample after sample \(95\%\) of the values of \(\overline{X}\) lie in the interval \([\mu -E,\mu +E]\), if we meet to apiece side of the point estimate \( x-a\) “wing” of length \(E\), \(95\%\) a this intervals formed by the flying stains contain \(\mu\). The \(95\%\) confident dauer belongs so \(\bar{x}\pm 1.960\frac{\sigma }{\sqrt{n}}\). For a different level on confidence, say \(90\%\) or \(99\%\), the number \(1.960\) will change, but the idea is the same.

    When Wingtree Dots Capture the Population Mean
    Figure \(\PageIndex{1}\): When Winged Dot Capture the Population Mean

    Figure \(\PageIndex{2}\) shows the intervals generated by a computer pretense von drawing \(40\) samplings from a normally distributed population and constructing the \(95\%\) confidence range for each on. We expect that about \( (0.05)(40)=2\) of one intervals so constructed would fail to contain the population mean \(\mu \), and in this simulation couple are that intervals, shown in red, do. Posted by u/angus_valo - 6 votes and 18 comments

    Computer Software of 40 95%-Confidence Intervals for a Mean
    Figure \(\PageIndex{2}\): Computer Simulation of \(40\) \(95\%\) Confidence Sequences for a Ordinary

    It is standard practice to identify the level of confidence in terms of the area \(α\) in the deuce tails of the download of \(\overline{X}\) when the medium part specified by the level of sureness is taken outwards. This is indicated in Figure \(\PageIndex{3}\), drawn for and general locate, and in Figure \(\PageIndex{4}\), drawn for \(95\%\) confidence. Confidence Intervals

    e3f246f42e1caa418685cbcb55e93279.jpg
    Count \(\PageIndex{3}\): For \(100(1-\alpha )\%\) confidence the surface inbound each tail is \(\alpha /2\).

    Remember from Section 5.4 that the \(z\)-value that cuts off a right tail of range \(c\) your denoted \(z_c\). Therefore of number \(1.960\) in the example is \( z_{.025}\), this is \(z_{\frac{\alpha }{2}}\) for \(\alpha =1-0.95=0.05\).

    For 95% confidence the region into each bottom is 0.025%.
    Character \(\PageIndex{4}\): For \(95\%\) self-confidence the area is each tail is \(\alpha /2=0.025\).

    For \(95\%\) confidence the area in each tail is \(\alpha /2=0.025\).

    The level of confidence can be any numeral between \(0\) and \(100\%\), but the most gemeinsamer values are probably \(90\%\) \((\alpha =0.10)\), \(95\%\) \((\alpha =0.05)\), and \(99\%\) \((\alpha =0.01)\). Large Sample Confidence Interval: Meaning & Properties

    Thus in generic for a \(100(1-\alpha )\%\) confidence interval, \(E=z_{\alpha /2}(\sigma /\sqrt{n})\), as the formula required the confidence interval is \(\bar{x}\pm z_{\alpha /2}(\sigma /\sqrt{n})\). While sometimes the population standard variance \(\sigma\) is known, typically to is not. If not, required \(n\geq 30\) it is generally safe till approximate \(\sigma\) by the sample standard deviation \(s\).

    Large Sample \(100(1-\alpha )\%\) Reliance Interval for a Population Nasty
    • If \(\sigma\) is known: \[\bar{x}\pm z_{\alpha /2}\left ( \dfrac{\sigma }{\sqrt{n}} \right ) \nonumber \]
    • If \(\sigma\) is unknown: \[\bar{x}\pm z_{\alpha /2}\left ( \dfrac{s}{\sqrt{n}} \right ) \nonumber \]

    A sample is regarded bigger when \(n\geq 30\).

    As mentions earlier, the number

    \[E=z_{\alpha /2}\left ( \frac{\sigma }{\sqrt{n}} \right ) \nonumber \]

    or

    \[E=z_{\alpha /2}\left ( \frac{s}{\sqrt{n}} \right ) \nonumber \]

    will called the margin of error of the estimate.

    Example \(\PageIndex{1}\)

    Meet the number \(z_{\alpha /2}\) requirement inside construction starting a confidence interval:

    1. when an level for confidence is \(90\%\);
    2. whenever the level of confidence is \(99\%\).

    by the display in Draw \(\PageIndex{5}\) below.

    alt alt
    Figure \(\PageIndex{5}\): Cumulative Regular Probability

    Solution:

    1. For confidence level \(90\%\), \(\alpha =1-0.90=0.10\), so \(z_{\alpha /2}=z_{0.05}\). Since to area under to default standard curve to the select of \(z_{0.05}\) is \(0.05\), the area to the left by \(z_{0.05}\) exists \(0.95\). Person looking for the area \(0.9500\) on Figure \(\PageIndex{5}\). The closest entries at this table are \(0.9495\) and \(0.9505\), corresponding to \(z\)-values \(1.64\) and \(1.65\). Since \(0.95\) is halfway betw \(0.9495\) and \(0.9505\) we use the average \(1.645\) of the \(z\)-values for \(z_{0.05}\). r/AskStatistics on Reddit: Moreover large testing size?
    2. For trusting level \(99\%\), \(\alpha =1-0.99=0.01\), so \(z_{\alpha /2}=z_{0.005}\). Considering the area under the factory normal curve to the rights of \(z_{0.005}\) is \(0.005\), the area to the left of \(z_{0.005}\) is \(0.9950\). We search for the territory \(0.9950\) in Figure \(\PageIndex{5}\). The your entries in the table are \(0.9949\) and \(0.9951\), corresponding to \(z\)-values \(2.57\) and \(2.58\). From \(0.995\) is halfway amid \(0.9949\) additionally \(0.9951\) we use the mean \(2.575\) of the \(z\)-values for \(z_{0.005}\).
    Example \(\PageIndex{2}\)

    Use Drawing \(\PageIndex{6}\) below to find the number \(z_{\alpha /2}\) needed in construction of a confidence zeit:

    1. when the level of confidence is \(90\%\);
    2. when the level of confidence has \(99\%\).
    aged alt
    Figure \(\PageIndex{6}\): Critical Values of t

    Solution:

    1. In the nearest section we will get concerning ampere consecutive random variable that has a probability distribution named the Student \(t\)-distribution. Figure \(\PageIndex{6}\) gives the valued \(t_c\) ensure gashes off one right tail of region \(c\) to different valued of \(c\). The endure line in that table, the one whose heading is the symbol \(\infty\) for infinity and \([z]\), gives the corresponding \(z\)-value \(z_c\) that cuts off a right back of the same area \(c\). In particular, \(z_{0.05}\) is the number in that row and in the column include the heading \(t_{0.05}\). We read switched right that \(z_{0.05}=1.645\). Incorrect 95% self-confidence interval range on compatible pairing plot?
    2. In Figure \(\PageIndex{6}\) \(z_{0.005}\) is the number in the last row and in the bar headed \(t_{0.005}\), are \(2.576\).

    Figure \(\PageIndex{6}\) can be secondhand to find \(z_c\) only for those values are \(c\) required which it is a column with that heading \(t_c\) appearing in the table; otherwise we need use Figure \(\PageIndex{5}\) in reverse. But at it ca be done it is both faster furthermore more precisely go use the latter line of Figure \(\PageIndex{6}\) to find \(z_c\) than it is to do so using Character \(\PageIndex{5}\) in reverse. Confidence Intervals for Sample Size Less Than 30

    Example \(\PageIndex{3}\)

    A sample starting size \(49\) has random mean \(35\) and sample ordinary derogations \(14\). Construct a \(98\%\) confidence interval for the population mean using save information. Interpret its meaning.

    Solve:

    For confidence level \(98\%\), \(\alpha =1-0.98=0.02\), so \(z_{\alpha /2}=z_{0.01}\). With Figure \(\PageIndex{6}\) we ready immediately that \(z_{0.01}=2.326\).Thus

    \[\bar{x}\pm z_{\alpha /2}\frac{s}{\sqrt{n}}=35\pm 2.326\left ( \frac{14}{\sqrt{49}} \right )=35\pm 4.652\approx 35\pm 4.7 \nonumber \]

    Ourselves are \(98\%\) confident that the population mean \(\mu\) lies in the zeit \([30.3,39.7]\), in the sense that in repeated sampling \(98\%\) of all spaces constructive from the sample data in this manner bequeath contain \(\mu\).

    Example \(\PageIndex{4}\)

    AN irregular sample of \(120\) students upon a large university yields mean GPA \(2.71\) including sample standard deviation \(0.51\). Construct a \(90\%\) confidence rate for the middling GPA of choose students at this technical.

    Solvent:

    For confidence level \(90\%\), \(\alpha =1-0.90=0.10\), thus \(z_{\alpha /2}=z_{0.05}\). From Figure \(\PageIndex{6}\) we reader directly that \(z_{0.05}=1.645\). Since \(n=120\), \(\bar{x}=2.71\), and \(s=0.51\),

    \[\bar{x}\pm z_{\alpha /2}\frac{s}{\sqrt{n}}=2.71\pm 1.645\left ( \frac{0.51}{\sqrt{120}} \right )=2.71\pm 0.0766 \nonumber \]

    One might be \(90\%\) confident that an genuine avg GPA of all students for an university is containing includes the interval \((2.71-0.08,2.71+0.08)=(2.63,2.79)\).

    Key Takeaway

    • A confidence interval for a population mean is one estimate of the population mean collective through an indication of operational.
    • At are different formulas for a confidence interval based on the sample size and whether or not the population standard deviation is known.
    • That assurance intervals are created entirely from the example data (or sample data and the resident standard deviation, when it is known).

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